The Mole. Chapter 2. Solutions for Practice Problems

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "The Mole. Chapter 2. Solutions for Practice Problems"

Transcription

1 Chapter 2 The Mole Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the full set of steps: Problem, What Is Required, What Is Given,, Act on Your Strategy, and. Where a shorter solution is appropriate, the Sample Problem contains only two steps: Problem and Solution. Where relevant, a Check Your Solution step is also included in the shorter Sample Problems. Solutions for Practice Problems Student Textbook page Problem The two stable isotopes of boron exist in the following proportions: 19.78% 11 (10.01 u) and 80.22% B (1.01 u). Calculate the average atomic mass of 2 boron. You need to find the average atomic mass of boron. You are given the relative abundance (proportions) and the atomic mass of each stable isotope. Express each isotopic proportion as a decimal and multiply this by its atomic mass. The average mass of boron is the sum of these products. Average atomic mass of B = u (0.1978) u (0.8022) = u Boron contains far more boron-11 than boron-10. Therefore, the atomic mass of the naturally occurring boron should be closer to u than to u. The calculated atomic mass matches this reasoning. A check in the periodic table confirms the value. 2. Problem In nature, silicon is composed of three isotopes. These isotopes (with their isotopic 28 abundances and atomic masses) are Si 29 (92.5, u), Si (4.67%, u), 30 and Si (3.10%, u). 14 Calculate the average mass of silicon. You need to find the atomic mass of silicon. You are given the relative abundance and the atomic mass of each isotope B 19

2 Express each isotopic proportion as a decimal and multiply this by its atomic mass. The average mass of silicon is the sum of these products. Average atomic mass of Si = (27.98 u)(0.92) + (28.97 u) (0.0467) + (29.97 u)(0.0310) = u Silicon contains far more silicon-28, than silicon-29 or silicon-30. Therefore, the atomic mass of the naturally occurring silicon should be closer to u than to u or u. The calculated atomic mass matches this reasoning and a check in the periodic table confirms the value. 3. Problem Copper is a corrosion-resistant metal that is used extensively in plumbing and wiring. 63 Copper exists as two naturally occurring isotopes: Cu 65 (62.93 u) and Cu (64.93 u). These isotopes have isotopic abundances of 69.1% and 30.9%, respectively. Calculate the average atomic mass of copper. You need to find the atomic mass of copper. You are given the relative abundance and the atomic mass of each isotope. Express each isotopic proportion as a decimal and multiply this by its atomic mass. The average mass of copper is the sum of these products. Average atomic mass of Cu = (62.93 u)(0.691) + (64.93 u)(0.309) = u Copper contains more copper-63 than copper-65. Therefore, the atomic mass of the naturally occurring copper should be closer to u than to u. The mean of the two isotopic masses is u. So, the calculated atomic mass of copper, which is lower than this mean value, is a reasonable answer. A check in the periodic table confirms the atomic mass calculated. 4. Problem Rubidium has as two isotopes: rubidum-85 (84.91 u) and rubidium-87 (86.91 u). If the average atomic mass of rubidium is u, determine the percentage abundance of each isotope. You have to find the percentage abundance of rubidium-85 and rubidium-87. The atomic mass of each isotope is given, as well as the average atomic mass of rubidium. First express the abundance of each isotope as a decimal instead of a percentage, giving a total abundance of both isotopes as 1. Let the abundance of be x and Rb 20

3 87 Rb 37 be 1 - x. Set up the equation and solve for x. Then multiply x by 100% to express it as percentage abundance. Average atomic mass = x( mass rubidium-85 ) + ( 1 - x) (mass rubidium-87) u = x(84.91) + ( 1 - x) (86.91) = x x 2x = x = 072. Therefore, the percentage abundance of rubidium is % = 72%. The percentage abundance of rubidium-87, then, is 28%. The mean of the atomic masses of the two isotopes is u. The average atomic mass of rubidium (85.47 u) is slightly less than this mean, and closer to the value of rubidium-85 (84.91 u). So you would expect rubidium to comprise of a greater abundance of rubidium-85, which matches the result. Solutions for Practice Problems Student Textbook pages Problem A small pin contains mol of iron, Fe. How many atoms of iron are in the pin? You need to find the number of atoms of iron in the pin. The pin contains mol of Fe. N A = atoms/mol. N = N A n The number of atoms would be the Avogadro constant multiplied by the number of moles of Fe in the pin. The number of atoms of Fe = mol ( atoms/mol) = atoms Therefore, there are iron atoms in the pin. Work backwards. Dividing the number of atoms by the Avogadro constant should give you the number of moles of Fe. 6. Problem A sample contains mol of gold, Au. How many atoms of gold are there in the sample? You need to find the number of atoms of gold in the sample. The sample contains mol of Au. N A = atoms/mol 21

4 N = N A n The number of atoms would be the Avogadro constant multiplied by the number of moles of gold in the sample. The number of atoms of Au = mol ( atoms/mol) = atoms Therefore, there are gold atoms in the sample. Work backwards. Dividing the number of atoms by the Avogadro constant should give you the number of moles of Au. 7. Problem How many formula units are contained in 0.21 mol of magnesium nitrate, Mg(NO 3 ) 2? You need to find the number of formula units in the magnesium nitrate. The sample has 0.21 mol. N A = formula units/mol N = N A n The number of formula units would be the Avogadro constant multiplied by the number of moles of Mg(NO 3 ) 2. The number of formula units of Mg(NO 3 ) 2 = 0.21 mol ( formula units/mol) = formula units Therefore, there are formula units of Mg(NO 3 ) 2 in the sample. Work backwards. Dividing the number of formula units by Avogadro constant should give you the number of moles of Mg(NO 3 ) Problem A litre of water contains 55.6 mol of water. How many molecules of water are in this sample? You need to find the number of molecules of water in the 1 L volume. There is 55.6 mol of water in the 1 L volume. N A = molecules/mol. N = N A n The number of molecules would be the Avogadro constant multiplied by the number of moles of water. The number of molecules of H 2 O = 55.6 mol ( molecules/mol) = molecules Therefore, there are water molecules in one litre of water. 22

5 Work backwards. Dividing the number of molecules by Avogadro s number should give you the number of moles of water. 9. Problem Ethyl acetate, C 4 H 8 O 2, is frequently used in nail polish remover. A typical bottle of nail polish remover contains about 2.5 mol of ethyl acetate. (a) How many molecules are in the bottle of nail polish remover? (b) How many atoms are in the bottle? (c) How many carbon atoms are in the bottle? (a) You need to find the number of molecules in the sample. (b) You need to find the number of atoms in the bottle. (c) You need to find the number of carbon atoms in the bottle. The sample consists of 2.5 mol. The molecular formula of ethyl acetate is C 4 H 8 O 2. The Avogadro constant N A = molecules/mol. (a) N = N A n The number of molecules would be the Avogadro constant multiplied by the number of moles of ethyl acetate. (b) The number of atoms per molecule, from the molecular formula of ethyl acetate, is 14. The total number of atoms in the bottle would be 14 multiplied by the number of molecules. (c) The number of carbon atoms in the molecular formula of ethyl acetate is 4. The total number of atoms in the bottle would be 4 multiplied by the number of molecules. (a) Number of molecules of C 4 H 8 O 2 = 2.5 mol ( molecules/mol) = molecules Therefore, there are molecules in the bottle. (b) The number of atoms = 14 atoms/molecule ( molecules) = atoms Therefore, there are atoms in the bottle. (c) The number of C atoms = 4 atoms/molecule ( molecules) = atoms Therefore, there are atoms of carbon in the bottle. (a) Work backwards. Dividing the number of molecules by Avogadro s number should give you the number of moles of ethyl acetate. (b) Work backwards. Dividing the number of atoms by the number of molecules in the bottle should give you the number of atoms in each molecule. (c) Work backwards. Dividing the number of carbon atoms in the bottle by the number of molecules in the bottle should give you the number of carbon atoms in each molecule. 10. Problem Consider a mol sample of sodium sulfate, Na 2 SO 4. (a) How many formula units are in the sample? (b) How many sodium ions, Na +, are in the sample?

6 (a) You need to find the number of formula units in the sample. (b) You need to find the number of sodium ions in the bottle. (c) How many sulfate ions, SO 4 2 (aq) are in the sample? (d) How many oxygen atoms are in the sample? The sample consists of mol. The molecular formula of sodium sulfate is Na 2 SO 4. N A = formula units/mol (a) The number of formula units would be the Avogadro constant multiplied by the number of moles of sodium sulfate. (b) There are 2 sodium ions per formula unit of sodium sulfate. The total number of sodium ions would be 2 multiplied by the number of formula units in the sample. 2 2 (c) There is one SO 4 (aq) ion per formula unit. The number of SO 4 (aq) ions will be the same as the number of formula units. (d) There are 4 sodium ions per formula unit. The total number of oxygen atoms will be 4 multiplied by the number of formula units. (a) Number of formula units = mol ( formula units/mol) = formula units Therefore, there are formula units in the sample. (b) The number of sodium ions = 2 ions/formula unit ( formula units) = sodium ions. Therefore, there are sodium ions in the sample. 2 (c) number of SO 4 (aq) ions = (d) number of oxygen atoms = = (a) Work backwards. Dividing the number of formula units by Avogadro s number should give you the number of mol of sodium sulfate. (b) Work backwards. Dividing the number of sodium ions by the number of formula units in the sample should give you the number of ions per formula unit. (c) Work backwards. Dividing the number of SO 4 2 (aq) ions by the number of formula units in the sample should give the number of ions per formula unit. (d) Work backwards. Dividing the number of oxygen atoms by the number of formula units in the sample should give the number of atoms per formula unit. 11. Problem A certain coil of pure copper wire has a mass of 5.00 g. This mass of copper corresponds to mol of copper. (a) How many atoms of copper are in the coil? (b) How many atoms of copper would there be in a coil of pure copper wire with a mass of g? (c) How many atoms of copper would there be in a coil of pure copper wire with a mass of 1.00 g? What Is Required? (a) You need to find the number of copper atoms in the coil. (b) You need to find the number of copper atoms in a coil with a mass of 10.0 g. (c) You need to find the number of copper atoms in a coil with a mass of 1.00 g. 24

7 What Is Given? m = 5.00 g n = mol (a) Use the formula N = n N A to determine the number of copper atoms in the coil. (b) Multiply your answer for (a) by 2. (c) Divide your answer for (b) by 10. atoms (a) N = mol mol = atoms (b) atoms = atoms 22 (c) atoms = atoms 10 (a) Work backwards. Dividing the number of atoms by Avogadro s number will give you the number of moles of copper. Solutions for Practice Problems Student Textbook page Problem A sample of bauxite ore contains molecules of aluminum oxide, Al 2 O 3. How many moles of aluminum oxide are in the sample? You need to find the number of moles in molecules of aluminum oxide. The number of molecules is N A = molecules/mol n = N N A Divide the number of molecules by the Avogadro constant to get the number of moles of aluminum oxide molecules Number of moles = molecules/mol = mol Therefore, there are 12.8 moles of aluminum oxide in the sample. Work backwards. Dividing the number of molecules by the number of moles should give you Avogadro s number. 13. Problem A vat of cleaning solution contains molecules of ammonia, NH 3. How many moles of ammonia are in the vat? You need to find the number of moles in molecules of ammonia. The number of molecules is N A = molecules/mol 25

8 Divide the number of molecules by Avogadro s number to get the number of moles of ammonia. 26 Number of moles molecules = ( molecules/mol) 3 = mol Therefore, there are mol of ammonia in the vat. Work backwards. Dividing the number of molecules by the number of moles should give you Avogadro s number. 14. Problem A sample of cyanic acid, HCN, contains atoms. How many moles of cyanic acid are in the sample? Hint: Find the number of molecules of HCN first. You need to find the number of moles of HCN in atoms of HCN. The number of atoms is The formula of cyanic acid is given. N A = molecules/mol From the formula, you know there are 3 atoms in every molecule of cyanic acid. The number of molecules of cyanic acid is the number of atoms divided by 3. Then, divide the number of molecules by the Avogadro constant to get the number of mol of HCN atoms Number of molecules of HCN = 3 atoms/molecule 22 = mol molecules Number of moles of HCN = ( molecules/mol) -2 = mol Therefore, there are mol or mol of cyanic acid in the sample. Work backwards. Dividing the number of molecules by the number of moles should give you the Avogadro constant. 15. Problem A sample of pure acetic acid, CH 3 COOH, contains 1.40 many moles of acetic acid are in the sample? 10 carbon atoms. How You need to find the number of moles of acetic acid that carries carbon atoms. The number of carbon atoms is The molecular formula of acetic acid is CH 3 COOH. N A = molecules/mol From the formula, you know there are 2 carbon atoms in every molecule of acetic acid. The number of molecules of acetic acid is the number of carbon atoms divided 26

9 by 2. Then, divide the number of molecules by the Avogadro constant to get the number of mol of acetic acid C atoms Number of molecules of acetic acid = 2 C - atoms/molecule Number of moles of acetic acid = molecules ( / molecules/mol) = mol Therefore, there are mol or mol of acetic acid in the sample. Work backwards. Dividing the number of molecules by the number of moles should give you the Avogadro constant. Solutions for Practice Problems Student Textbook page Problem State the molar mass of each element. (a) xenon, Xe (b) osmium, Os (c) barium, Ba (d) tellurium, Te You have to find the molar mass of the elements listed. The chemical symbols and chemical names are given. With no other information given, the only source of the answer is a periodic table. The average atomic mass of the element is numerically equivalent to its molar mass. From the average atomic mass in the periodic table, the numerically equivalent molar masses are: (a) Xe: g/mol (b) Os: 190. g/mol (c) Ba: g/mol (d) Te: g/mol Periodic table values can vary in some publications, depending on the number of significant digits recorded. Check with other books or online sources to verify the most commonly accepted atomic mass units (and hence the molar mass) of the elements. 17. Problem Find the molar mass of each compound. (a) ammonia, NH 3 (b) glucose, C 6 H 12 O 6 (c) potassium dichromate, K 2 Cr 2 O 7 (d) iron(iii) sulfate, Fe 2 (SO 4 ) 3 = You have to find the molar masses of the compounds listed. 22 molecules 27

10 The chemical formulas of the compounds are given. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. The molar masses of the compounds are: (a) NH 3 : g/mol + 3(1.01 g/mol) = g/mol (b) C 6 H 12 O 6 : 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol) = g/mol (c) K 2 Cr 2 O 7 : 2(39.10 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol) = g/mol (d) Fe 2 (SO 4 ) 3 : 2(55.85 g/mol) + 3[32.07 g/mol + 4(16.00 g/mol)] = g/mol Using round numbers for a quick check, for example in the case of (a), = This estimate is close to the answer of g/mol. 18. Problem Strontium may be found in nature as celestite, SrSO 4. Find the molar mass of celestite. You have to find the molar mass of celestite. The chemical formula of celestite is SrSO 4. The average atomic mass can be obtained from any Periodic Table. The molar mass of celestite is the sum of all the molar masses of its component elements. Find the atomic mass of 1 mol of Sr, 1 mol of S, and 4 mol of O from the Periodic Table and add them together. Molar mass of SrSO 4 = g/mol g/mol + 4(16.00 g/mol) = g/mol Using round numbers for a quick check, you get (4 16) = 183. This estimate is close to the answer of g/mol. 19. Problem What is the molar mass of the ion [Cu(NH 3 ) 4 ] 2+? You have to find the molar mass of the [Cu(NH 3 ) 4 ] 2+ ion. The chemical formula of the ion is [Cu(NH 3 ) 4 ] 2+. The average atomic mass can be obtained from any Periodic Table. The molar mass of the ion is the sum of all the molar masses of its component elements. Find the atomic mass of 1 mol of Cu, 4 mol of N, and 12 mol of H from the Periodic Table and add them together. The mass of the two missing electrons from each ion may be neglected since the mass of an electron is extremely small compared to the mass of an atom. 28

11 Molar mass of [Cu(NH 3 ) 4 ] 2+ = g/mol + 4(14.01 g/mol) + 12(1.01 g/mol) = g/mol g/mol g/mol = g/mol Using round numbers for a quick check, you get = 131. This estimate is close to the answer of g/mol. Solutions for Practice Problems Student Textbook page Problem Calculate the mass of each molar quantity. (a) 3.90 mol of carbon, C (b) 2.50 mol of ozone, O 3 (c) mol of propanol, C 3 H 8 O (d) mol of ammonium dichromate, (NH 4 ) 2 Cr 2 O 7 You need to find the mass of the substances listed. The number of moles of the substances are given as well as the chemical formulas. The molar mass of the elements that make up the substances can be found in a Periodic Table. In order to convert moles to grams, you need to determine the molar mass (M) of the substance. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m) of the substance. (a) For C: M = g/mol m = 3.90 mol g/mol = 46.8 g (b) For O 3 : M = 3(16.00 g/mol) = g/mol m = 2.50 mol g/mol = 120 g (c) For C 3 H 8 O:M = 3(12.01 g/mol) + 8(1.01 g/mol) g/mol = g/mol m = mol g/mol = g (d) For (NH 4 ) 2 Cr 2 O 7 : M = 2[14.01 g/mol + 4(1.01 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol) = g/mol m = mol g/mol = g 29

12 Work backwards. Divide the calculated mass by the number of moles given. It should g give the molar mass of the substance. For example, in (a), = 12.0 g/mol. This matches the molar mass of carbon mol 21. Problem For each group, which sample has the largest mass? (a) 5.00 mol of C, 1.50 mol of Cl 2, 0.50 mol of C 6 H 12 O 6 (b) 7.31 mol of O 2, 5.64 mol of CH 3 OH, 12.1 mol of H 2 O You need to find the substance with the largest mass among the substances listed. The number of moles of the substances are given as well as the chemical formulas. The molar mass of the elements that make up the substances can be found in a Periodic Table. In order to convert moles to grams, you need to determine the molar mass (M) of the substance. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m) of the substance. (a) For C: M = g/mol m = 5.00 mol g/mol = g For Cl 2 : M = g/mol m = 1.50 mol g/mol = g For C 6 H 12 O 6 : M = g/mol m = 0.50 mol g/mol = g Therefore, the sample of Cl 2 has the largest mass. (b) For O 2 : M = g/mol; m = 7.31 mol g/mol = 4 g For CH 3 OH: M = g/mol; m = 5.64 mol g/mol = 181 g For H 2 O: M = g/mol; m = 12.1 mol g/mol = 218 g Therefore, the sample of O 2 has the largest mass. Work backward to check the mass. Divide the calculated mass by the number of moles given. It should give the molar mass of the substance. For example, for C in (a), 60.6 g =12.0 g/mol. This matches the molar mass of carbon mol 22. Problem A litre, 1000 ml, of water contains 55.6 mol. What is the mass of a litre of water. You need to find the mass of 1000 ml of water. The number of moles of water is given as well as the volume. The molar mass of the elements that make up water can be found in a Periodic Table. 30

13 In order to convert moles to grams, you need to determine the molar mass (M) of water. Find the atomic mass (which is numerically equivalent to the molar mass) for 2 mol H and 1 mol O from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m). Molar mass = 2(1.01 g/mol) g/mol = g/mol m = 55.6 mol g/mol = g = g Since the density of water is 1.0 g/ml, the mass of 1000 ml is 1000 g or 1 kg. This agrees with the result calculated by the mole method.. Problem To carry out a particular reaction, a chemical engineer needs 255 mol of styrene, C 8 H 8. How many kilograms of styrene does the engineer need? You need to find the mass of styrene needed. The number of moles of the styrene is given as well as the chemical formula. The molar mass of the elements that make up styrene can be found in a Periodic Table. In order to convert moles to grams, you need to determine the molar mass (M) of the styrene. The molar mass of a compound is the sum of all the molar masses of its component elements. Find the atomic mass (which is numerically equivalent to the molar mass) of the elements from the periodic table and add them together. Then multiply the molar mass by the given number of moles to obtain the mass (m) of the substance. Divide the mass in gram by 10 3 g/kg for a kilogram conversion. M = 8(12.01 g/mol) + 8(1.01 g/mol) = g/mol m = 255 mol g/mol = g Mass in kilograms = g/10 3 g/kg = kg Therefore, the engineer needs 26.6 kg of styrene for the reaction. Work backwards. Divide the calculated mass by the number of moles given. It should g give the molar mass of the substance. = 104 g mol, which agrees with 255 mol the calculations. Solutions for Practice Problems Student Textbook page Problem Calculate the number of moles in each sample. (a) 103 g of Mo (b) g of Pd (c) kg of Cr (d) 56.3 mg of Ge You need to find the number of moles in the given mass of elements listed. 31

14 The masses of the elements are given. You have access to a periodic table. In each case, determine the molar mass (M) from a periodic table and divide it into the given sample mass (m) to obtain the number of moles (n). Use n = m. In the M case of (c) and (d), first convert the given amount to grams before proceeding with the calculation. Act On Your Strategy (a) For Mo: M = g/mol; m = 103 g 103 g n = = 107. mol g/mol (b) For Pd: M = g/mol; m = g g n = = 124 mol g/mol (c) For Cr: M = g/mol; m = kg 10 3 g/kg = 736 kg 736 g n = = mol g/mol (d) For Ge: M = g/mol; m = 56.3 mg 103 mg/g = g g - n = = mol g/mol In each case, work backward to reestablish the given mass of the element. For example, in (a), 1.07mol g/mol = 103 g. This value matches the amount given in the question. 25. Problem How many moles of compound are in each sample? (a) 39.2 g of silicon dioxide, SiO 2 (b) 7.34 g of nitrous acid, HNO 2 (c) kg of carbon tetrafluoride, CF 4 (d) mg of 1-iodo-2,3,-dimethylbenzene, C 8 H 9 I You need to find the number of moles in the given mass of substances listed. The masses of the substances are given as well as the chemical formulas. You have access to a periodic table. In each case, determine the molar mass (M) from a periodic table and divide it into the given sample mass (m) to obtain the number of moles (n). Use n = m. In the M case of (c) and (d), first convert the given amount to grams before proceeding with the calculation. Act On Your Strategy (a) For SiO 2 : M = g/mol; m = 39.2 g 39.2 g n = = mol g/mol (b) For HNO 2 : M = g/mol; m = 7.34 g 7.34 g n = = mol g/mol 32

15 3 8 (c) For CF 4 : M = g/mol; m = ( g 10 kg) 10 g/kg = g g n = 10 3 g/kg = mol g/mol (d) For C 8 H 9 I: M = 2.07 g/mol; m = mg/10 3 mg/g = g g -8 n = = mol 2.07 g/mol In each case, work backward to reestablish the given mass of the element. For example, in (a), mol g/mol = 39.2 g. This value matches the amount given in the question. 26. Problem Sodium chloride, NaCl, can be used to melt snow. How many moles of sodium chloride are in a 10 kg bag? You need to find the number of moles of NaCl in the bag. The mass of the salt is 10 kg. You have access to a Periodic Table. Determine the molar mass (M) from a Periodic Table and divide it into the given sample mass (m) to obtain the number of moles (n). Use n = m. Remember to convert the given amount to grams before proceeding with the calculation. M Act On Your Strategy M = g/mol; m = 10 kg 10 3 g/kg = g 10,000 g -2 n = = mol g/mol There is mol of salt in 10 kg. Work backward to reestablish the given mass of the salt. ( mol) g/mol = g. This value closely matches the amount given in the question. 27. Problem Octane, C 8 H 18, is a principal ingredient of gasoline. Calculate the number of moles in a 20.0 kg sample of octane. You need to find the number of moles of octane in the sample of gasoline. The mass of the octane is 20.0 kg and its chemical formula is given. You have access to a Periodic Table. Determine the molar mass (M) of octane using the atomic mass units from a periodic table. Divide M into the given sample mass (m) to obtain the number of moles (n). Use n = m. Remember to convert the given amount to grams before proceeding M with the calculation. Act On Your Strategy M = g/mol; m = 20.0 kg 10 3 g/kg = g; 20,000 g n = = 175 mol g/mol 5 33

16 Work backward to reestablish the given mass of octane. 175 mol g/mol = g or 20.0 kg. This value closely matches the 20.0 kg amount given in the question. Solutions for Practice Problems Student Textbook page Problem Determine the mass of each sample. (a) formula units of ZnCl 2 (b) formula units of Pb 3 (PO 4 ) 2 (c) molecules of C 15 H 21 N 3 O 15 (d) molecules of N 2 O 5 You need to find the masses of the substances listed. You are given the number of formula units or molecules of the substances. N A = formula units/mol Convert the number of formula units or the number of molecules into moles (n) by dividing by the Avogadro constant. Then multiply the number of moles by the molar mass (M) to get the mass in grams (m) of the substance. 24 ( formula units) (a) n = = mol ( ) formula units / mol M of ZnCl 2 = g / mol; 3 Therefore, m = mol g / mol = g or g 21 (b) n = formula units = mol ( ) formula units / mol M of Pb 3 (PO 4 ) 2 = g/mol; Therefore, m = 0.01 mol g/mol = 9.98 g (c) ( molecules) n = = 151. mol ( ) molecules / mol M of C 15 H 21 N 3 O 15 = g/mol; Therefore, m = 1.51 mol g/mol = g 29 (d) ( molecules) 5 n = = mol ( ) molecules / mol M of N 2 O 5 = g/mol; Therefore, m = 1.99 mol 10 5 g/mol g/mol = 9.98 g or kg. Work backwards. Dividing the mass by the number of moles should give the molar mass of the substance. For example, in (a), 10.0 mol = 136. The results match. 29. Problem What is the mass of lithium in 254 formula units of lithium chloride, LiCl? You need to find the mass of lithium in the LiCl. 34

17 You are given the number of formula units of lithium chloride as well as the chemical formula. N A = formula units/mol. From the chemical formula, you know there is 1 unit of lithium for every formula unit of lithium chloride, so you can assume that there are 254 units of lithium in the sample. Convert the number of units of the lithium into moles (n) by dividing by the Avogadro constant. Then multiply the number of moles by the molar mass (M) of lithium to get the mass in grams (m) of the lithium. n = 254 formula units -22 = mol ( ) formula units / mol M of Li = 6.94 g/mol; Therefore, m = ( ) mol 6.94 g/mol = g. Work backwards. Dividing the mass by the number of moles should give the molar mass of lithium. = 6.94 g/mol. The results match g mol 30. Problem A sample of potassium oxide, K 2 O, contains K + ions. What is the mass of the sample? You must find the mass of potassium oxide in that contains K + ions. You know the number of K + ions and the Avogadro constant is ions/mol. Convert the number of ions to moles (n) by dividing by the Avogadro constant. From the formula of potassium oxide, K 2 O, one mol of K 2 O contains 2 potassium ions. Therefore the moles of K 2 O must be half the number of moles of K +. Divide the moles of K + by 2. Change this number of moles of K 2 O to grams by multiplying by the molar mass of K 2 O. M of K 2 O is 94.2 g/mol ion n = mol K = ions ions/mol mol of K 2 O = 3.31 mol Therefore m = 3.31 mol K 2 O 94.2 g/mol = 312 g Work backwards. 312 g K 2O 94.2 g/mol + 2 mol K ion/ mol = ions 1 mol K O This is the number of ions given in the problem Problem A sample of ammonium chloride, NH 4 Cl, contains hydrogen atoms. What is the mass of the sample? 35

18 You must find the mass of ammonium chloride in that contains hydrogen atoms. You know the number of hydrogen atoms and the Avogadro constant is ions/mol. Convert the number of atoms to moles (n) by dividing the Avogadro constant. From the formula of ammonium chloride, NH 4 Cl, one mol of NH 4 Cl contains 4 hydrogen 1 atoms. Therefore, the moles of NH 4 Cl must be the number of moles of hydrogen 4 atoms. Divide the number of moles of hydrogen atoms by 4. Change this number of moles of NH 4 Cl to grams by multiplying by the molar mass of NH 4 Cl. M of NH 4 Cl is 53.5 g/mol atom n = =1.50 mol hydrogen atoms atoms/mol mol of NH 4 Cl = mol Therefore m = mol NH 4 Cl 53.5 g/mol = 20.1 g NH 4 Cl Work backwards g NH4Cl 53.5 g/mol 4 mol H atom atom/mol 1 mol NH Cl 4 = atoms This closely matches the number of hydrogen atoms given in the problem. 32. Problem Express the mass of a single atom of titanium, Ti, in grams. You have to find the mass of a single Ti atom. The chemical symbols and chemical names are given. You have access to a periodic table, and the Avogadro constant, N A = atoms/mol. According to the Avogadro constant, there are atoms of Ti in every mole of titanium. Therefore, the number of moles (n) of 1 atom of Ti is 1 mol divided by atoms. Multiply this value by the molar mass (M) of Ti to obtain the mass (m) of 1 atom of Ti. Number of moles of 1 atom = 1 mol/ atoms = mol M of Ti = g/mol Therefore, m = mol g/mol = g. Work backwards. Dividing the mass by the number of moles should give the molar mass of titanium. g. The results match = g/mol.. - mol 36

19 33. Problem Vitamin B 2, C 17 H 20 N 4 O 6, is also called riboflavin. What is the mass, in grams, of a single molecule of riboflavin? You have to find the mass of a single riboflavin molecule. The chemical formula of the vitamin is given. You have access to a periodic table, and the Avogadro constant, N A = molecules/mol. According to the Avogadro constant, there are molecules of riboflavin in every mole of the vitamin. Therefore, the number of moles (n) of 1 molecule of riboflavin is 1 mol divided by molecules. Multiply this value by the molar mass (M) of C 17 H 20 N 4 O 6 to obtain the mass (m) of a single molecule. 1mol Number of moles of 1 molecule = = mol molecules M of C 17 H 20 N 4 O 6 = g/m; Therefore, m = mol g/mol = g. Work backwards. Dividing the mass by the number of moles should give the molar mass of titanium. 22 g g/mol. The results match. Solutions for Practice Problems Student Textbook page Problem Determine the number of molecules or formula units in each sample. (a) 10.0 g of water, H 2 O (b) 52.4 g of methanol, CH 3 OH (c).5 g of disulfur dichloride, S 2 Cl 2 (d) g of lead(ii) phosphate, Pb 3 (PO 4 ) 2 You need to find the number of formula units or molecules of the substances listed. You are given the masses of the substances as well as their molecular formulas. N A = formula units/mol First convert the mass (m) to moles (n), using the molar mass (M) of the substance and the formula n = m/m. Multiplying this value by the Avogadro constant will yield the number of formula units (or the number of molecules) g g/mol (a) n = = mol Therefore, the number of molecules of water = mol ( ) molecules/mol = molecules. (b) n = 52.4 g g/mol = mol = 1.63 mol Therefore, the number of molecules of CH 3 OH. = 1.63 mol ( ) molecules/mol = molecules. 37

20 .5 g g/mol (c) n = = mol Therefore, the number of molecules of S 2 Cl 2 = mol ( ) molecules/mol = molecules. (d) n = g g/mol = mol Therefore, the number of formula units of Pb 3 (PO 4 ) 2 = mol ( ) formula units/mol = formula units. In each case, work backwards to reestablish the mass given in the question. For example, in (a), each mole of water has a mass of g. Therefore, molecules of water has a mass of: molecules ( 1 mol ) ( g ) molecules 1 mol = 9.99 g The answer is close to the 10.0 g given in the question (a). Your answer is reasonable. 35. Problem How many atoms of hydrogen are in molecules of sodium glutamate, NaC 5 H 8 NO 4? You need to find the number of H atoms in the given number of molecules. You are given the number of molecules of sodium glutamate as well as its molecular formula. From the formula, you can tell that there are 8 hydrogen atoms for every formula unit of sodium glutamate. The total number of hydrogen atoms is therefore 8 multiplied by the given number of molecules. Total number of H atoms = 8 atoms/molecule ( ) molecules = atoms Using round numbers for a quick check, = or This estimate is close to the answer. 36. Problem How many molecules are in a 64.3 mg sample of tetraphosphorus decoxide, P 4 O 10? You need to find the number of molecules in the P 4 O 10 sample. You are given the mass of the tetraphosphorus decoxide as well as its molecular formula. You know the Avogadro constant, N A = molecules/mol. First convert the mass (m) to moles (n), using the molar mass (M) of the tetraphosphorus decoxide and the formula n = m/m. Multiplying this value by Avogadro s number will yield the number of molecules. Remember to convert the mass to grams before proceeding with the calculation. m = mg = g; M = g/mol 3 10 mg/g 38

21 Therefore, n = g = mol g/mol Hence, the number of molecules of P 4 O 10 = mol ( ) molecules/mol = molecules. Work backwards to reestablish the mass given in the question. Therefore, molecules of P 4 O 10 has a mass of: molecules ( 1 mol molecules ) ( g 1 mol = g or 64.1 mg. The answer is close to the 64.3 mg given in the question. 37. Problem (a) How many formula units are in a g sample of potassium chlorate, KClO 3? (b) How many ions (chlorate and potassium) are in this sample? (a) You need to find the number of formula units in the given sample. (b) You need to find the number of chlorate and potassium ions in this sample. (a) You are given the mass of the potassium chlorate. (b) You are given its molecular formula. You know Avogadro s number, N A = formula units/mol. (a) First convert the mass (m) to moles (n), using the molar mass (M) of the potassium chlorate and the formula n = m. Multiplying this value by the Avogadro M constant will yield the number of formula units. (b) From the chemical formula of potassium chlorate, you know there is 1 ion of potassium and 1 ion of chlorate for every formula unit of potassium chlorate. The total number of ions will therefore be the number of formula units calculated in (a) multiplied by 2. (a) m = g; M = g/mol; Therefore, n = g = mol g/mol Hence, the number of formula units of KClO 3 = mol ( ) molecules/mol = formula units (b) Each formula unit of KClO 3 consists of two ions, so there are 2 ions/formula unit ( ) formula units = ions in the sample. (a) Work backwards to reestablish the mass given in the question. Therefore, formula units of KClO 3 has a mass of: formula units ( 1 mol molecules ) ) ( g 1 mol = g. The answer is close to the g given in the question. (b) Using round numbers for a quick check, 2 ( ) = This estimate is close to the answer. ) 39

22 Solutions for Practice Problems Student Textbook page Problem A balloon contains 2.0 L of helium gas at STP. How many moles of helium are present? What Is Required? You need to find the number of moles of helium at STP. What Is Given? The gas is at STP, so its molar volume = 22.4 L/mol. That is, V i = 22.4 L, n i = 1 mol, V f = 2.0 L. Use Avogadro s law to solve for n f. n i V i = n f 1mol 2.0 L 22.4 L \ n V f = = mol i Therefore, the number of moles of helium is mol. The molar volume is 22.4 L/mol. Working with the numbers, 2.0 L = 22 L/mol mol The results match. 39. Problem How many moles of gas are present in 11.2 L at STP? How many molecules? What Is Required? (a) You need to find the number of moles of gas at the given volume at STP. (b) You need to find the number of molecules of gas at STP. What Is Given? (a) At STP, V i = 22.4 L; n i = 1 mol; V f = 11.2 L (b) You know the Avogadro constant = molecules/mol. (a) Use Avogadro s law to solve for n f. (b) Multiply n f by the Avogadro constant to obtain the number of molecules at this given volume. n i V i = n f 1mol 11.2 L 22.4 L \ n V f = = mol f Number of molecules of gas = mol ( ) molecules/mol = molecules The molar volume is 22.4 L/mol. Working with the numbers, 11.2 L mol = 22.4 L/mol. The results match. 40. Problem What is the volume, at STP, of 3.45 mol of argon gas? What Is Required? You need to find the volume of the argon gas for the given number of moles. What Is Given? At STP, V i = 22.4 L; n i = 1 mol; n f = 3.45 mol 40

23 Use Avogadro s law to solve for V f. n i V i = n f V f \ V f = 3.45 mol 22.4 L 1.00 mol = 77.3 L The molar volume is 22.4 L/mol. Working with the numbers, 77.3 L The results match mol = 22.4 L/mol. 41. Problem At STP, a container holds g of nitrogen gas, g of oxygen gas, g of carbon dioxide gas, and g of ammonia gas. What is the volume of the container? What Is Required? You need to find the volume of the container. What Is Given? At STP, V i = 22.4 L; n i = 1 mol The mass of each gas is given. You have access to the periodic table for the molar masses of the elements. The volume of the container will comprise the total number of moles of all the gases present. Divide each mass of gas by its own molar mass to obtain its number of moles. Add the number of moles together, and then use this as n f in Avogadro s law to solve for V f g n nitrogen = = mol n oxygen = n carbon dioxide = n ammonia = g/mol g g/mol g g/mol g g/mol = mol = 1.50 mol = 1.00 mol Total number of moles = mol mol mol mol = 3.50 mol Applying Avogadro s law: n i V i = n f 3.50 mol 22.4 L 1.00 mol \ V V f = = 78.4 L f Therefore, the volume of the container is 78.4 L. The molar volume at STP is 22.4 L/mol. Working with the numbers, = 22.4 L/mol. The results match L 3.50 mol 42. Problem (a) What volume do 2.50 mol of oxygen occupy at STP? (b) How many molecules are present in this volume of oxygen? (c) How many oxygen atoms are present in this volume of oxygen? What Is Required? (a) You need to find the volume of oxygen from the given number of moles. (b) You need to find the number of molecules of this oxygen. (c) You need to find the number of atoms present in this volume of oxygen. 41

24 What Is Given? At STP, V i = 22.4 L; n i = 1 mol; n f = 2.50 mol You know Avogadro s number is molecules/mol. You know oxygen is a diatomic gas. (a) Use Avogadro s law to solve for V f. (b) Multiply V f by the Avogadro number to obtain the number of molecules of oxygen gas. (c) Multiply the number of molecules of oxygen gas by 2 atoms to obtain the number of oxygen atoms in the sample. n i 2.50 mol 22.4 L (a) = n f \ V V i V f = = 56.0 L f 1.00 L (b) Number of molecules of oxygen = 250. mol ( ) molecules/ mol = molecules (c) There are two oxygen atoms in one molecule of oxygen gas. Therefore, the number of oxygen atoms = 2( ) molecules = atoms. The molar volume is 22.4 L/mol. Working with the numbers, 56.0 L = 22.4 L/mol mol The results match. 43. Problem What volume do atoms of neon occupy at STP? What Is Required? You need to find the volume of the gas for the given number of moles. What Is Given? At STP, V i = 22.4 L; n i = 1 mol Number of atoms = Avogadro s number is molecules/mol. Neon is a monatomic gas. Neon is a monatomic so the number of atoms will equal the number of molecules of gas. Multiply the number of molecules by Avogadro s number to get the number of moles of gas as n f. Use Avogadro s law and n f to solve for V f Number of moles of Ne = 24 molecules = 3.32 mol n i V i = n f 3.32 mol 22.4 L 1.00 mol molecules/mol \ V V f = = 74.4 L f Therefore, the volume of the atoms of neon at STP is 74.4 L. The molar volume is 22.4 L/mol. Working with the numbers, 74.4 L = 22.4 L/mol mol The results match. 42

Counting Atoms and Molecules: The Mole

Counting Atoms and Molecules: The Mole Chapter 5 Practice Problems Counting Atoms and Molecules: The Mole Solutions for Practice Problems Student Textbook pages 167 168 1. Problem The two stable isotopes of boron exist in the following proportions:

More information

SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001

SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, Chemistry 11, McGraw-Hill Ryerson, 2001 SCH 4C1 Unit 2 Problem Set Questions taken from Frank Mustoe et all, "Chemistry 11", McGraw-Hill Ryerson, 2001 1. A small pin contains 0.0178 mol of iron. How many atoms of iron are in the pin? 2. A sample

More information

Exploring Gas Laws. Chapter 12. Solutions for Practice Problems. Student Textbook page 477

Exploring Gas Laws. Chapter 12. Solutions for Practice Problems. Student Textbook page 477 Chapter 12 Exploring Gas Laws Solutions for Practice Problems Student Textbook page 477 1. Problem At 19 C and 100 kpa, 0.021 mol of oxygen gas, O 2(g), occupy a volume of 0.50 L. What is the molar volume

More information

Study Guide For Chapter 7

Study Guide For Chapter 7 Name: Class: Date: ID: A Study Guide For Chapter 7 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The number of atoms in a mole of any pure substance

More information

Chapter 3 Mass Relationships in Chemical Reactions

Chapter 3 Mass Relationships in Chemical Reactions Chapter 3 Mass Relationships in Chemical Reactions Student: 1. An atom of bromine has a mass about four times greater than that of an atom of neon. Which choice makes the correct comparison of the relative

More information

INTRODUCTORY CHEMISTRY Concepts and Critical Thinking

INTRODUCTORY CHEMISTRY Concepts and Critical Thinking INTRODUCTORY CHEMISTRY Concepts and Critical Thinking Sixth Edition by Charles H. Corwin Chapter 9 The Mole Concept by Christopher Hamaker 2011 Pearson Education, Inc. Chapter 9 1 Avogadro s Number Avogadro

More information

Unit 2. Molar Mass Worksheet

Unit 2. Molar Mass Worksheet Unit 2 Molar Mass Worksheet Calculate the molar masses of the following chemicals: 1) Cl 2 8) UF 6 2) KOH 9) SO 2 3) BeCl 2 10) H 3 PO 4 4) FeCl 3 11) (NH 4 ) 2 SO 4 5) BF 3 12) CH 3 COOH 6) CCl 2 F 2

More information

Solution. Practice Exercise. Concept Exercise

Solution. Practice Exercise. Concept Exercise Example Exercise 9.1 Atomic Mass and Avogadro s Number Refer to the atomic masses in the periodic table inside the front cover of this textbook. State the mass of Avogadro s number of atoms for each of

More information

W1 WORKSHOP ON STOICHIOMETRY

W1 WORKSHOP ON STOICHIOMETRY INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of

More information

B. Elements: We cannot determine how many electrons are lost for the elements b/c their in their valence electrons can change.

B. Elements: We cannot determine how many electrons are lost for the elements b/c their in their valence electrons can change. Unit 6 Notepack: Chapters 9 &10 Chemical Quantities 9.1 Naming Ions NAME Period: A. ions: Ions made of single. B. Elements: There is a pattern in predicting how many electrons are lost and gained for the

More information

Chemical Quantities: The Mole Chapter 7 Assignment & Problem Set

Chemical Quantities: The Mole Chapter 7 Assignment & Problem Set Chemical Quantities: The Mole Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. Chemical Quantities: The Mole 2 Study Guide: Things You Must Know

More information

Chemical Proportions in Compounds

Chemical Proportions in Compounds Chapter 3 Chemical Proportions in Compounds Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem

More information

CHEMICAL QUANTITIES. Chapter 10

CHEMICAL QUANTITIES. Chapter 10 CHEMICAL QUANTITIES Chapter 10 What is a mole? A unit of measurement in chemistry 1 mole of a substance = 6.02 x 10 23 (Avagadro s number) representative particles of a substance Representative particle

More information

PART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points)

PART I: MULTIPLE CHOICE (30 multiple choice questions. Each multiple choice question is worth 2 points) CHEMISTRY 123-07 Midterm #1 Answer key October 14, 2010 Statistics: Average: 74 p (74%); Highest: 97 p (95%); Lowest: 33 p (33%) Number of students performing at or above average: 67 (57%) Number of students

More information

CHAPTER 3 COMPOUNDS AND MOLECULES

CHAPTER 3 COMPOUNDS AND MOLECULES Chapter 3 Compounds and Molecules Page 1 CHAPTER 3 COMPOUNDS AND MOLECULES 3-1. Octane, a component of gasoline, has eight carbon atoms and eighteen hydrogen atoms per molecule. Its formula is written

More information

1. Balance the following equation. What is the sum of the coefficients of the reactants and products?

1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1 Fe 2 O 3 (s) + _3 C(s) 2 Fe(s) + _3 CO(g) a) 5 b) 6 c) 7 d) 8 e) 9 2. Which of the following equations

More information

Problem Solving. Mole Concept

Problem Solving. Mole Concept Skills Worksheet Problem Solving Mole Concept Suppose you want to carry out a reaction that requires combining one atom of iron with one atom of sulfur. How much iron should you use? How much sulfur? When

More information

Chapter 4 Chemical Composition

Chapter 4 Chemical Composition Chapter 4 Chemical Composition 4.1 (a) mole; (b) Avogadro s number; (c) empirical formula; (d) solute; (e) molarity; (f) concentrated solution 4. (a) molar mass; (b) percent composition by mass; (c) solvent;

More information

Moles, Molecules, and Grams Worksheet Answer Key

Moles, Molecules, and Grams Worksheet Answer Key Moles, Molecules, and Grams Worksheet Answer Key 1) How many are there in 24 grams of FeF 3? 1.28 x 10 23 2) How many are there in 450 grams of Na 2 SO 4? 1.91 x 10 24 3) How many grams are there in 2.3

More information

1. In 1928, 3.3 g of a new element was isolated from 660 kg of the ore molybdenite. The percent by mass of this element in the ore was:

1. In 1928, 3.3 g of a new element was isolated from 660 kg of the ore molybdenite. The percent by mass of this element in the ore was: AP Chemistry Practice Test #1 - Key Chapters 1, 2, and 3 1. In 1928, 3.3 g of a new element was isolated from 660 kg of the ore molybdenite. The percent by mass of this element in the ore was: a. 5 % b.

More information

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations CHE11 Chapter Chapter Stoichiometry: Calculations with Chemical Formulas and Equations 1. When the following equation is balanced, the coefficients are. NH (g) + O (g) NO (g) + H O (g) (a). 1, 1, 1, 1

More information

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296)

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296) 10 CHEMICAL QUANTITIES SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296) This section defines the mole and explains how the mole is used to measure matter. It also teaches you how to calculate

More information

MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles]

MOLE CONVERSION PROBLEMS. 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] MOLE CONVERSION PROBLEMS 1. What is the molar mass of MgO? [40.31 g/mol] 2. How many moles are present in 34 grams of Cu(OH) 2? [0.35 moles] 3. How many moles are present in 2.5 x 10 23 molecules of CH

More information

Element of same atomic number, but different atomic mass o Example: Hydrogen

Element of same atomic number, but different atomic mass o Example: Hydrogen Atomic mass: p + = protons; e - = electrons; n 0 = neutrons p + + n 0 = atomic mass o For carbon-12, 6p + + 6n 0 = atomic mass of 12.0 o For chlorine-35, 17p + + 18n 0 = atomic mass of 35.0 atomic mass

More information

Calculating Atoms, Ions, or Molecules Using Moles

Calculating Atoms, Ions, or Molecules Using Moles TEKS REVIEW 8B Calculating Atoms, Ions, or Molecules Using Moles TEKS 8B READINESS Use the mole concept to calculate the number of atoms, ions, or molecules in a sample TEKS_TXT of material. Vocabulary

More information

Chapter 7. Bellringer. Table of Contents. Chapter 7. Chapter 7. Objectives. Avogadro s Number and the Mole. Chapter 7. Chapter 7

Chapter 7. Bellringer. Table of Contents. Chapter 7. Chapter 7. Objectives. Avogadro s Number and the Mole. Chapter 7. Chapter 7 The Mole and Chemical Table of Contents Chemical Formulas Bellringer List as many common counting units as you can. Determine how many groups of each unit in your list are present in each of the following

More information

CHAPTER. Copyright by Holt, Rinehart and Winston. All rights reserved.

CHAPTER. Copyright by Holt, Rinehart and Winston. All rights reserved. CHAPTER 222 Galaxies have hundreds of billions of stars. The universe may have as many as sextillion stars that s 1000 000 000 000 000 000 000 (or 1 10 21 ) stars. Such a number is called astronomical

More information

U3-LM2B-WS Molar Mass and Conversions

U3-LM2B-WS Molar Mass and Conversions U3-LM2B-WS Molar Mass and Conversions Name: KEY 1. The molar mass of chlorine is: 2 x 35.45 g/mol Cl = 70.90 g/mol Cl 2 (Remember that chlorine exists as a diatomic molecule in nature) 2. The molar mass

More information

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1

Chapter 3. Stoichiometry: Ratios of Combination. Insert picture from First page of chapter. Copyright McGraw-Hill 2009 1 Chapter 3 Insert picture from First page of chapter Stoichiometry: Ratios of Combination Copyright McGraw-Hill 2009 1 3.1 Molecular and Formula Masses Molecular mass - (molecular weight) The mass in amu

More information

Molar Mass Worksheet Answer Key

Molar Mass Worksheet Answer Key Molar Mass Worksheet Answer Key Calculate the molar masses of the following chemicals: 1) Cl 2 71 g/mol 2) KOH 56.1 g/mol 3) BeCl 2 80 g/mol 4) FeCl 3 162.3 g/mol 5) BF 3 67.8 g/mol 6) CCl 2 F 2 121 g/mol

More information

0.786 mol carbon dioxide to grams g lithium carbonate to mol

0.786 mol carbon dioxide to grams g lithium carbonate to mol 1 2 Convert: 2.54 x 10 22 atoms of Cr to mol 4.32 mol NaCl to grams 0.786 mol carbon dioxide to grams 2.67 g lithium carbonate to mol 1.000 atom of C 12 to grams 3 Convert: 2.54 x 10 22 atoms of Cr to

More information

21 st Century Chemistry Multiple Choice Question in Topic 3 Metals Unit 11

21 st Century Chemistry Multiple Choice Question in Topic 3 Metals Unit 11 21 st Century Chemistry Multiple Choice Question in Topic 3 Metals Unit 11 1. Consider the equation: 2Ca(s) + O 2 (g) 2CaO(s) Which of the following statements are correct? (1) Calcium and oxygen are reactants.

More information

Atomic mass and the mole

Atomic mass and the mole Atomic mass and the mole An equation for a chemical reaction can provide us with a lot of useful information. It tells us what the reactants and the products are in the reaction, and it also tells us the

More information

Mole Notes.notebook. October 29, 2014

Mole Notes.notebook. October 29, 2014 1 2 How do chemists count atoms/formula units/molecules? How do we go from the atomic scale to the scale of everyday measurements (macroscopic scale)? The gateway is the mole! But before we get to the

More information

4.4 Calculations Involving the Mole Concept

4.4 Calculations Involving the Mole Concept 44 Section 43 Questions 1 Define Avogadro s constant, and explain its significance in quantitative analysis 2 Distinguish between the terms atomic mass and molar mass 3 Calculate the mass of a molecule

More information

4. Magnesium has three natural isotopes with the following masses and natural abundances:

4. Magnesium has three natural isotopes with the following masses and natural abundances: Exercise #1 Atomic Masses 1. The average mass of pennies minted after 1982 is 2.50 g and the average mass of pennies minted before 1982 is 3.00 g. In a sample that contains 90.0% new and 10.0% old pennies,

More information

Chemical Proportions in Compounds

Chemical Proportions in Compounds Chapter 6 Chemical Proportions in Compounds Solutions for Practice Problems Student Textbook page 201 1. Problem A sample of a compound is analyzed and found to contain 0.90 g of calcium and 1.60 g of

More information

= 11.0 g (assuming 100 washers is exact).

= 11.0 g (assuming 100 washers is exact). CHAPTER 8 1. 100 washers 0.110 g 1 washer 100. g 1 washer 0.110 g = 11.0 g (assuming 100 washers is exact). = 909 washers 2. The empirical formula is CFH from the structure given. The empirical formula

More information

ATOMS. Multiple Choice Questions

ATOMS. Multiple Choice Questions Chapter 3 ATOMS AND MOLECULES Multiple Choice Questions 1. Which of the following correctly represents 360 g of water? (i) 2 moles of H 2 0 (ii) 20 moles of water (iii) 6.022 10 23 molecules of water (iv)

More information

Test 3 Formula and math of formula review. 2. The oxide of metal X has the formula XO. Which group in the Periodic Table contains metal X?

Test 3 Formula and math of formula review. 2. The oxide of metal X has the formula XO. Which group in the Periodic Table contains metal X? Name: Sunday, November 04, 2007 Test 3 Formula and math of formula review 1. Which is an example of a binary compound? 1. acetic acid 3. potassium hydroxide 2. nitric acid 4. potassium oxide 2. The oxide

More information

MOLECULAR MASS AND FORMULA MASS

MOLECULAR MASS AND FORMULA MASS 1 MOLECULAR MASS AND FORMULA MASS Molecular mass = sum of the atomic weights of all atoms in the molecule. Formula mass = sum of the atomic weights of all atoms in the formula unit. 2 MOLECULAR MASS AND

More information

neutrons are present?

neutrons are present? AP Chem Summer Assignment Worksheet #1 Atomic Structure 1. a) For the ion 39 K +, state how many electrons, how many protons, and how many 19 neutrons are present? b) Which of these particles has the smallest

More information

Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test

Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test Unit 3 Notepack Chapter 7 Chemical Quantities Qualifier for Test NAME Section 7.1 The Mole: A Measurement of Matter A. What is a mole? 1. Chemistry is a quantitative science. What does this term mean?

More information

Suggested Resources Textbook: Chapter 6 (6.3, 6.4, 6.5 & 6.6)

Suggested Resources Textbook: Chapter 6 (6.3, 6.4, 6.5 & 6.6) The Mole Big Picture Ideas: 1. The mole is a unit of count. 2. Using conversion factors, one can convert between mass, moles, particles and volume for a given substance. 3. Empirical data including percent

More information

602X10 21 602,000,000,000, 000,000,000,000 6.02X10 23. Pre- AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements

602X10 21 602,000,000,000, 000,000,000,000 6.02X10 23. Pre- AP Chemistry Chemical Quan44es: The Mole. Diatomic Elements Pre- AP Chemistry Chemical Quan44es: The Mole Mole SI unit of measurement that measures the amount of substance. A substance exists as representa9ve par9cles. Representa9ve par9cles can be atoms, molecules,

More information

Unit 10A Stoichiometry Notes

Unit 10A Stoichiometry Notes Unit 10A Stoichiometry Notes Stoichiometry is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations

More information

Chapter 3. Stoichiometry of Formulas and Equations

Chapter 3. Stoichiometry of Formulas and Equations Chapter 3 Stoichiometry of Formulas and Equations Chapter 3 Outline: Mole - Mass Relationships in Chemical Systems 3.1 The Mole 3.2 Determining the Formula of an Unknown Compound 3.3 Writing and Balancing

More information

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS

CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS FOLLOW UP PROBLEMS 3.1A Plan: The mass of carbon must be changed from mg to g. The molar mass of carbon can then be used to determine the number of moles.

More information

Chapter 3 Mass Relations in Chemistry; Stoichiometry

Chapter 3 Mass Relations in Chemistry; Stoichiometry Chapter 3 Mass Relations in Chemistry; Stoichiometry MULTIPLE CHOICE 1. An atomic mass unit (amu) is defined as a. 1.60 10-19 C. b. the mass of 1 mole of hydrogen-s. c. the mass of 1 hydrogen-. d. the

More information

STOICHIOMETRY STOICHIOMETRY. Measurements in Chemical Reactions. Mole-Mole Relationships. Mass-Mass Problem. Mole-Mole Relationships

STOICHIOMETRY STOICHIOMETRY. Measurements in Chemical Reactions. Mole-Mole Relationships. Mass-Mass Problem. Mole-Mole Relationships STOICHIOMETRY STOICHIOMETRY The analysis of the quantities of substances in a chemical reaction. Stoichiometric calculations depend on the MOLE- MOLE relationships of substances. Measurements in Chemical

More information

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296)

Name Date Class CHEMICAL QUANTITIES. SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296) Name Date Class 10 CHEMICAL QUANTITIES SECTION 10.1 THE MOLE: A MEASUREMENT OF MATTER (pages 287 296) This section defines the mole and explains how the mole is used to measure matter. It also teaches

More information

Chapter 7 Part II: Chemical Formulas and Equations. Mr. Chumbley Chemistry 1-2

Chapter 7 Part II: Chemical Formulas and Equations. Mr. Chumbley Chemistry 1-2 Chapter 7 Part II: Chemical Formulas and Equations Mr. Chumbley Chemistry 1-2 SECTION 3: USING CHEMICAL FORMULAS Molecules and Formula Unit We have not yet discussed the different ways in which chemical

More information

Sample Exercise 2.1 Illustrating the Size of an Atom

Sample Exercise 2.1 Illustrating the Size of an Atom Sample Exercise 2.1 Illustrating the Size of an Atom The diameter of a US penny is 19 mm. The diameter of a silver atom, by comparison, is only 2.88 Å. How many silver atoms could be arranged side by side

More information

Chapter 1: Moles and equations. Learning outcomes. you should be able to:

Chapter 1: Moles and equations. Learning outcomes. you should be able to: Chapter 1: Moles and equations 1 Learning outcomes you should be able to: define and use the terms: relative atomic mass, isotopic mass and formula mass based on the 12 C scale perform calculations, including

More information

TOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights.

TOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights. TOPIC 7. CHEMICAL CALCULATIONS I - atomic and formula weights. Atomic structure revisited. In Topic 2, atoms were described as ranging from the simplest atom, H, containing a single proton and usually

More information

Chemistry Post-Enrolment Worksheet

Chemistry Post-Enrolment Worksheet Name: Chemistry Post-Enrolment Worksheet The purpose of this worksheet is to get you to recap some of the fundamental concepts that you studied at GCSE and introduce some of the concepts that will be part

More information

MODERN CHEMISTRY. 8. What is the formula for aluminum sulfate? 9. What is the formula for barium hydroxide?

MODERN CHEMISTRY. 8. What is the formula for aluminum sulfate? 9. What is the formula for barium hydroxide? MODERN CHEMISTRY Date Name Period Multiple Choice - Identify the choice that best completes the statement or answers the question. Use answer-key worksheet (Page 6) to key in answers to questions. 1. A

More information

The Nature of Chemistry

The Nature of Chemistry CHAPTER 1 The Nature of Chemistry Objectives You will be able to do the following. 1. Describe how science in general is done. 2. Given a description of a property of a substance, identify the property

More information

Periodic Table, Valency and Formula

Periodic Table, Valency and Formula Periodic Table, Valency and Formula Origins of the Periodic Table Mendelѐѐv in 1869 proposed that a relationship existed between the chemical properties of elements and their atomic masses. He noticed

More information

The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis

The Mole Concept. A. Atomic Masses and Avogadro s Hypothesis The Mole Concept A. Atomic Masses and Avogadro s Hypothesis 1. We have learned that compounds are made up of two or more different elements and that elements are composed of atoms. Therefore, compounds

More information

b. N 2 H 4 c. aluminum oxalate d. acetic acid e. arsenic PART 2: MOLAR MASS 2. Determine the molar mass for each of the following. a. ZnI 2 b.

b. N 2 H 4 c. aluminum oxalate d. acetic acid e. arsenic PART 2: MOLAR MASS 2. Determine the molar mass for each of the following. a. ZnI 2 b. CHEMISTRY DISCOVER UNIT 5 LOTS OF PRACTICE ON USING THE MOLE!!! PART 1: ATOMIC MASS, FORMULA MASS, OR MOLECULAR MASS 1. Determine the atomic mass, formula mass, or molecular mass for each of the following

More information

SCH3UI-02 Final Examination Review (Fall 2014)

SCH3UI-02 Final Examination Review (Fall 2014) SCH3UI-02 Final Examination Review (Fall 2014) 1. Chlorine has an atomic number of 17. Create a Bohr Rutherford diagram of a Cl-35 atom. To achieve a stable arrangement, is this atom most likely to gain

More information

TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas.

TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas. TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas. Percentage composition of elements in compounds. In Topic 1 it was stated that a given compound always has the same composition by

More information

Chemical Composition Review Mole Calculations Percent Composition. Copyright Cengage Learning. All rights reserved. 8 1

Chemical Composition Review Mole Calculations Percent Composition. Copyright Cengage Learning. All rights reserved. 8 1 Chemical Composition Review Mole Calculations Percent Composition Copyright Cengage Learning. All rights reserved. 8 1 QUESTION Suppose you work in a hardware store and a customer wants to purchase 500

More information

Molecular Formula: Example

Molecular Formula: Example Molecular Formula: Example A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? 1 CHAPTER 3 Chemical

More information

MOLES AND MOLE CALCULATIONS

MOLES AND MOLE CALCULATIONS 35 MOLES ND MOLE CLCULTIONS INTRODUCTION The purpose of this section is to present some methods for calculating both how much of each reactant is used in a chemical reaction, and how much of each product

More information

Chemical calculations

Chemical calculations Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg +

More information

Ch. 10 The Mole I. Molar Conversions

Ch. 10 The Mole I. Molar Conversions Ch. 10 The Mole I. Molar Conversions I II III IV A. What is the Mole? A counting number (like a dozen) Avogadro s number (N A ) 1 mole = 6.022 10 23 representative particles B. Mole/Particle Conversions

More information

1. How many hydrogen atoms are in 1.00 g of hydrogen?

1. How many hydrogen atoms are in 1.00 g of hydrogen? MOLES AND CALCULATIONS USING THE MOLE CONCEPT INTRODUCTORY TERMS A. What is an amu? 1.66 x 10-24 g B. We need a conversion to the macroscopic world. 1. How many hydrogen atoms are in 1.00 g of hydrogen?

More information

Solution : 22.4 dm 3 of a gas at STP 1 mol of gas molar mass of the gas Now, 4.48 dm 3 of NH 3 at STP 3.49 g dm 3 of NH 3 at STP.

Solution : 22.4 dm 3 of a gas at STP 1 mol of gas molar mass of the gas Now, 4.48 dm 3 of NH 3 at STP 3.49 g dm 3 of NH 3 at STP. Formulae 1. One mole of atoms Mass of elements Atomic mass Atomic mass 2. Mass of one atom 3 6.022 10 2 3. Number of moles (n) Mass of substance Molar mass of substance 4. Number of molecules n Avogadro

More information

Chemical Calculations: Formula Masses, Moles, and Chemical Equations

Chemical Calculations: Formula Masses, Moles, and Chemical Equations Chemical Calculations: Formula Masses, Moles, and Chemical Equations Atomic Mass & Formula Mass Recall from Chapter Three that the average mass of an atom of a given element can be found on the periodic

More information

Labs: Chalk Lab Bubble Gum Lab Mole Lab the Works. Activities: Mole Airlines Flight Ebola Mole. Quizzes:

Labs: Chalk Lab Bubble Gum Lab Mole Lab the Works. Activities: Mole Airlines Flight Ebola Mole. Quizzes: The Mole **Make sure you get your daily work signed off on. That way, when we test you'll have the grade you earned instead of Videos freaking out about 50's in the book. #1 -- The Mole #2 -- Particles

More information

More on ions (Chapters 2.1 and )

More on ions (Chapters 2.1 and ) More on ions (Chapters 2.1 and 3.5 3.7) Ion: an atom or molecule that has a net electrical charge. Examples: Na + (sodium ion), Cl - (chloride), NH 4 + (ammonium). Anion: a negative ion, formed when electrons

More information

Chapter 8 Chemical Quantities

Chapter 8 Chemical Quantities Chapter 8 Chemical Quantities Introductory Info The atomic masses of the elements on the periodic table are in the units. These measurements are based on the mass of the standard isotope of the element,

More information

Page Which element is a noble gas? (1) krypton (3) antimony (2) chlorine (4) manganese

Page Which element is a noble gas? (1) krypton (3) antimony (2) chlorine (4) manganese 1. Which characteristics describe most nonmetals in the solid phase? (1) They are malleable and have metallic luster. (2) They are malleable and lack metallic luster. (3) They are brittle and have metallic

More information

Programme in mole calculation

Programme in mole calculation Programme in mole calculation Step 1 Start at the very beginning Atoms are very small indeed! If we draw a line 1 metre long, 6,000,000,000 (6 billion) atoms could be lined end to end. So a scientist cannot

More information

1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10

1. P 2 O 5 2. P 5 O 2 3. P 10 O 4 4. P 4 O 10 Teacher: Mr. gerraputa Print Close Name: 1. A chemical formula is an expression used to represent 1. mixtures, only 3. compounds, only 2. elements, only 4. compounds and elements 2. What is the total number

More information

Description of the Mole Concept:

Description of the Mole Concept: Description of the Mole Concept: Suppose you were sent into the store to buy 36 eggs. When you picked them up you would get 3 boxes, each containing 12 eggs. You just used a mathematical device, called

More information

Topic 4 National Chemistry Summary Notes. Formulae, Equations, Balancing Equations and The Mole

Topic 4 National Chemistry Summary Notes. Formulae, Equations, Balancing Equations and The Mole Topic 4 National Chemistry Summary Notes Formulae, Equations, Balancing Equations and The Mole LI 1 The chemical formula of a covalent molecular compound tells us the number of atoms of each element present

More information

Ch. 6 Chemical Composition and Stoichiometry

Ch. 6 Chemical Composition and Stoichiometry Ch. 6 Chemical Composition and Stoichiometry The Mole Concept [6.2, 6.3] Conversions between g mol atoms [6.3, 6.4, 6.5] Mass Percent [6.6, 6.7] Empirical and Molecular Formula [6.8, 6.9] Bring your calculators!

More information

A dozen. Molar Mass. Mass of atoms

A dozen. Molar Mass. Mass of atoms A dozen Molar Mass Science 10 is a number of objects. A dozen eggs, a dozen cars, and a dozen people are all 12 objects. But a dozen cars has a much greater mass than a dozen eggs because the mass of each

More information

B) atomic number C) both the solid and the liquid phase D) Au C) Sn, Si, C A) metal C) O, S, Se C) In D) tin D) methane D) bismuth B) Group 2 metal

B) atomic number C) both the solid and the liquid phase D) Au C) Sn, Si, C A) metal C) O, S, Se C) In D) tin D) methane D) bismuth B) Group 2 metal 1. The elements on the Periodic Table are arranged in order of increasing A) atomic mass B) atomic number C) molar mass D) oxidation number 2. Which list of elements consists of a metal, a metalloid, and

More information

Practice questions for Chapter 8

Practice questions for Chapter 8 Practice questions for Chapter 8 2) How many atoms of nickel equal a mass of 58.69 g? (Refer to the Periodic Table.) A) 1 B) 28 C) 58.69 D) 59 E) 6.02 x 1023 Answer: E Section: 8.1 Avogadro's Number 6)

More information

B. chemical equation - set of symbols and numbers which represent a chemical change. Reactants Products

B. chemical equation - set of symbols and numbers which represent a chemical change. Reactants Products 1 I. Introduction A. chemical reaction any chemical change B. chemical equation set of symbols and numbers which represent a chemical change 1. reactants what you begin with 2. products what you end up

More information

MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY

MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY MASS RELATIONS IN CHEMISTRY; STOICHIOMETRY [MH5; Ch. 3] Each element has its own unique mass. The mass of each element is found on the Periodic Table under the chemical symbol for the element (it is usually

More information

CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT

CHAPTER 3 Calculations with Chemical Formulas and Equations. atoms in a FORMULA UNIT CHAPTER 3 Calculations with Chemical Formulas and Equations MOLECULAR WEIGHT (M. W.) Sum of the Atomic Weights of all atoms in a MOLECULE of a substance. FORMULA WEIGHT (F. W.) Sum of the atomic Weights

More information

STOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to:

STOICHIOMETRY UNIT 1 LEARNING OUTCOMES. At the end of this unit students will be expected to: STOICHIOMETRY LEARNING OUTCOMES At the end of this unit students will be expected to: UNIT 1 THE MOLE AND MOLAR MASS define molar mass and perform mole-mass inter-conversions for pure substances explain

More information

The Mole IT IS JUST A STANDARD NUMBER OF ATOMS/MOLECULES

The Mole IT IS JUST A STANDARD NUMBER OF ATOMS/MOLECULES The Mole S 6.02 X 10 23 IT IS JUST A STANDARD NUMBER OF ATOMS/MOLECULES Relative Masses To understand relative scales, let s s ignore electrons and compare atoms by total number of nuclear particles. Hydrogen,

More information

CHAPTER 8: CHEMICAL COMPOSITION

CHAPTER 8: CHEMICAL COMPOSITION CHAPTER 8: CHEMICAL COMPOSITION Active Learning: 1-4, 6-8, 12, 18-25; End-of-Chapter Problems: 3-4, 9-82, 84-85, 87-92, 94-104, 107-109, 111, 113, 119, 125-126 8.2 ATOMIC MASSES: COUNTING ATOMS BY WEIGHING

More information

TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas.

TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas. TOPIC 8. CHEMICAL CALCULATIONS II: % composition, empirical formulas. Percentage composition of elements in compounds. In Topic 1 it was stated that a given compound always has the same composition by

More information

2 The Structure of Atoms

2 The Structure of Atoms CHAPTER 4 2 The Structure of Atoms SECTION Atoms KEY IDEAS As you read this section, keep these questions in mind: What do atoms of the same element have in common? What are isotopes? How is an element

More information

CHEM 10113, Exam 1 September 19, 2006

CHEM 10113, Exam 1 September 19, 2006 CHEM 10113, Exam 1 September 19, 2006 Name (please print) All answers must use the correct number of significant figures, and must show units! 1. (15 points) Mercury poisoning is a debilitating disease

More information

Mole Concept and Stoichiometry

Mole Concept and Stoichiometry 1 Mole Concept and Stoichiometry Concept Mole Concept Introduction This is our common experience that when we go to market to buy something, a few things we always get in definite numbers. For example,

More information

Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O

Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Balance the following equation: KClO 3 + C 12 H 22 O 11 KCl + CO 2 + H 2 O Ans: 8 KClO 3 + C 12 H 22 O 11 8 KCl + 12 CO 2 + 11 H 2 O 3.2 Chemical Symbols at Different levels Chemical symbols represent

More information

Unit 9 Stoichiometry Notes (The Mole Continues)

Unit 9 Stoichiometry Notes (The Mole Continues) Unit 9 Stoichiometry Notes (The Mole Continues) is a big word for a process that chemist s use to calculate amounts in reactions. It makes use of the coefficient ratio set up by balanced reaction equations

More information

midterm1, 2009 Name: Class: Date:

midterm1, 2009 Name: Class: Date: Class: Date: midterm1, 2009 Record your name on the top of this exam and on the scantron form. Record the test ID letter in the top right box of the scantron form. Record all of your answers on the scantron

More information

Chem. Ch. 10 ~ THE MOLE NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics.

Chem. Ch. 10 ~ THE MOLE NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. Chem. Ch. 10 ~ THE MOLE NOTE: Vocabulary terms are in boldface and underlined. Supporting details are in italics. 10.1 Notes I. Measuring Matter A. SI unit of chemical quantity = the mole (abbreviated

More information

4. Magnesium has three natural isotopes with the following masses and natural abundances:

4. Magnesium has three natural isotopes with the following masses and natural abundances: Exercise #1 Atomic Masses 1. The average mass of pennies minted after 1982 is 2.50 g and the average mass of pennies minted before 1982 is 3.00 g. In a sample that contains 90.0% new and 10.0% old pennies,

More information

Quantities in Chemical Reactions

Quantities in Chemical Reactions Chapter Quantities in Chemical Reactions Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains

More information

AP Chemistry Prep - Summer Assignment 2013

AP Chemistry Prep - Summer Assignment 2013 AP Chemistry Prep - Summer Assignment 2013 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which value has only 4 significant digits? a. 6.930 c. 8450

More information