Sample Problem: Find the percentage composition of MgCO 3.

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1 Chemistry H: Form TR4.5A TEST 4 REVIEW Name Date Period Test Review # 4 Formula Mass. The masses of ionic and covalent compounds are found the same way from the formula. The atomic masses of the elements in the compound and the formula are used to determine the mass. The mass determined from the formula is called a formula mass. A molecular mass is a type of formula mass. The terms are sometimes used interchangeably. Formula masses are determined by following the steps in the box to the right. The results are in atomic mass units (amu) Empirical Formulas. The chemical formula for a molecular compound shows the number and type of atoms present in a molecule. Ionic crystals are a collections of ions. The chemical formula for an ionic compound shows the ratio ions in the compound. The ratio of ions in the formula for an ionic compound is always in lowest terms. A chemical formula in which the ratio of the elements are in lowest terms is called an empirical formula. The molecular formula for glucose (C 6 H 1 O 6 ) is not an empirical formula. All the subscripts are divisible by six. When the subscripts are divided by six, the empirical formula for glucose, CH O, is obtained. Some molecular formulas, such as the one for carbon dioxide, CO, are already empirical formulas without being reduced. There are two skills you need to learn in order to work with empirical formulas: Finding the empirical formula from the molecular formula; and finding the molecular formula from the empirical formula and the molecular mass. To find the empirical formula from the molecular A compound with an empirical formula of CHO has a molecular mass of 90 amu. What is its molecular formula? Step 1: Step : Step : Determine the empirical formula mass. CH O C = 1 1 = 1 H = 1 = O = 16 1 = 16 0 Divide the molecular mass by the empirical formula mass to determine the multiple. 90 = 0 Multiply the empirical formula by the by the multiple to find the molecular formula [CH O] = C H 6 O Finding the Formula Mass Find the formula mass of CuSO4 Step 1: Step : Step : Element Look up the mass of each element on the Periodic Table and round it off. Multiply each element s atomic mass by its subscript to get the product. Add the products together to get the total Atomic Mass Subscript Product Cu 64 1 = 64 S 1 = O 16 4 = 64 TOTAL 160 formula, divide all the subscripts by the greatest common factor. To find the molecular formula from the empirical formula and the molecular mass. Percent Composition. Percentage composition is determined by finding the formula mass of a compound, multiplying the mass of each element by 100, and dividing the product by the formula mass of the compound. Use the periodic table to find the masses of individual elements. See the below. : Find the percentage composition of MgCO. Formula Mass Percentage Composition Mg = 4 1 = 4 % Mg = = 9 C = 1 1 = 1 % C = = 14 O =16 =48 % O = = Chemical Equations. Chemical equations provide a shorthand way to easily describe what occurs during a chemical reaction. In a typical chemical equation, the reactants are written on the left, while the products are written on the right. The reactants and products are separated by an arrow, or yield sign, which indicates that reactants yield products. ( REACTANTS ÿ PRODUCTS ) There are other symbols as well that show the state of the chemicals involved in the reaction. They are: (s) or 9 for a solid precipitate; (R) for a liquid; (g) or 8 for a gas; and (aq) for dissolved in water or aqueous. Symbols can also be used to show other factors involved in the reaction such as sources of energy used. These include: for heat or # for light. These symbols are written above or below the yield sign because they are neither reactants nor products. The complete equation shows the identity of the reactants and products using chemical formulas and symbols, the phases of the reactants and products, any energy changes involved in the reaction, and the mole ratios of all the substances indicated by the coefficients. Equations may occasionally be written omitting information

2 TEST 4 REVIEW Page about phases or energy changes. The example below shows a complete chemical equation with all the components. KClO (s) MnO ( s) Substance Formula Mass Gram Formula Mass carbon 1 amu 1 g sodium chloride (NaCl) KCl(s) + O (g) In the above reaction, the equation shows that the reactant is solid potassium chlorate, the products are solid potassium chloride and oxygen gas, manganese dioxide is a catalyst, and the reaction is endothermic. Symbols for manganese dioxide and heat are shown above and below the yield sign because they are neither reactants nor products. Reaction Types. Chemical reactions can be grouped into four basic types. They are direct combination or synthesis, decomposition, single replacement or substitution, and double replacement or exchange of ions. An example of synthesis is shown below: catalyst N( g) + H( g) NH( g) Synthesis often results in the formation of only one product from two reactants, but not always. Combustion, as in the following example, CH 4 (g) + O (g) ÿ CO (g) + H O, is also a form of synthesis because the oxygen combines with both the metal and the nonmetal to form two oxides. Decomposition is the reverse of synthesis. One reactant breaks apart to form several products. This is what happens when hydrogen peroxide decomposes over time to leave behind plain, ordinary water [H O (aq) ÿ H O(R) + O (g)]. During a single replacement reaction, a more active metal replaces a less active metal in a compound, or a more active nonmetal replaces a less active nonmetal in a compound. This is what happens when a metal becomes corroded by an acid [Fe(s) + 6HCl(aq) ÿ FeCl (aq) + H (g)]. In single replacement reactions, an element is reacting with a compound. Double replacement reactions occur between aqueous compounds. The cations and 58 amu 58 g glucose (C 6 H 1 O 6 ) 180 amu 180 g Patterns of the Reaction Types Legend: < A and C = metals < B and D = nonmetals Direct combination (synthesis) A + B ÿ AB or AB + D ÿ AD + BD Decomposition AB ÿ A + B Single Replacement (substitution) AB + C ÿ CB + A or AB + D ÿ AD + B Double Replacement (Exchange of Ions) AB + CD ÿ AD + CB anions switch partners. If an insoluble precipitate forms, the reaction is an end reaction, otherwise the result is an aqueous mixture of ions. An example of a double replacement reaction is AgNO (aq) + NaCl(aq) ÿ NaNO (aq) + AgCl(s). Conservation of Mass. Matter is neither created nor destroyed. During a chemical reaction the mass does not change. A properly written equation shows conservation of mass. Balancing the equation will make it show conservation of mass. Balancing Equations. The equation at the top of the box to the right does not show conservation of mass. Starting with two molecules of hydrogen, as shown in the equation at the bottom of the box by writing a coefficient in front of the hydrogen and forming two molecules of water by writing a coefficient in front of the water shows conservation. Coefficients are used to balance equations. Coefficients make the number of atoms of each type the same on the reactant and product side. As a result, coefficients make the mass the same on the reactant and product side of the equation. Balancing is done by counting the number and type of atoms on the reactant and product side of the equation and making them equal. Moles. A mole is a formula mass expressed in grams. (1 mole = 1 gram formula mass). Atomic mass units are too small to measure on a laboratory balance, but grams are not. An atom of carbon has a mass of 1 amu and a molecule of glucose has a mass of 180 amu. Each mass represents one particle. Since the mass ratios in formula masses and gram formula masses are the same (1 amu:180 amu::1 g:180 g), the ratio of particles must still be the same (1mole:1 mole). The gram formula mass (GFM) is the number of grams in 1 mole. This results in the mathematical relationships shown above and to the right. H + O ÿ H O + 18 H + O ÿ H O () + = (18) 6 = 6 g 1. GFM = mole. g = GFM mole g. mole = GFM

3 TEST 4 REVIEW Page How many moles of oxygen are consumed when 0.6 moles of hydrogen burns to produce water? Step 1: W r i t e a balanced equation and determine the mole ratios from the equation Step : Identify the known and the unknown Step : S e t u p a proportion and solve for the unknown mole ratio moles H (g) + O (g) ÿ H O 1 known 0.6 unknown x 1 = 06. mol x x = 06. mol x = 0. mol Stoichiometry. Stoichiometry is the branch of chemistry that deals with the application of the laws of definite proportions and of the conservation of mass and energy to chemical activity. It shows the quantitative relationship between constituents of a chemical reaction. Stoichiometric calculations are based on several assumptions. It is assumed that the reaction has no side reactions, the reaction goes to completion, and the reactants are completely consumed. One type of problem that can be solved stoichiometrically is based on the mole ratios of a balanced equation. A sample problem is shown to the left. Find the molecular formula for a compound composed of 5.9% hydrogen and 94.1% oxygen and having a molecular weight of 4 amu. Step 1: Assume a 100 g sample Step : Find the mass of each element in the sample mass of H = 5.9 % of 100 g = 5.9 g mass of O = 94.1 % of 100 g = 94.1 g Formulas from Masses. The molecular formula for a compound can be determined from the percentage composition by assuming the sample has a mass of 100 g. Using the percentages, the number of grams out of 100 can be determined for each component. This can be converted to moles by dividing by the GFM. The mole ratio and empirical formula can be determined by dividing each number of moles by the smallest number of moles. The atomic masses are added together to find the empirical formula mass. The empirical formula mass is divided into the molecular weight to find the number of times n, the formula is repeated. Finally, n is multiplied by the empirical formula to find the molecular formula. See the to the right. Mass/Mass Problems. With a balanced equation, a Periodic Table, and some knowledge of chemistry, you can figure out how much of any product will form from a given amount of reactant. There is a sample problem below solved by the factor label method. You will notice that, in applying the factor label method, you are first converting grams of the known to moles, then moles of the known to moles of the unknown using a proportion from the coefficients of the balanced equation, and, How much oxygen is needed to produce 7.0 g of water by burning hydrogen? Step : Convert grams to moles moles of H 59. g = 1gmol = 5.9 moles moles of O 941. g = 16 gmol = 5.9 moles Step 4: Find the mole ratio by dividing both numbers by the smaller number = 1 H = 1 O empirical formula = HO Step 5: Find the empirical formula mass atomic mass of H = 1 atomic mass of O = 16 EFM = 17 Step 6: Find the number of times, n, the empirical formula is repeated and multiply through M.W. = n = 4 = EFM 17 molecular formula (HO) n = (HO) = H O finally, moles of the unknown to grams as shown above. You can use the equations to the lower right instead of using the factor label method. Step 1: Step : Step : 7g HO Write a balanced equation H + O ÿ H O Calculate the GFM of the known and unknown. O H O O = 16 = H = 1 = O =16 1 = Apply the factor label method 1molHO 1mol g O O = 4g 18g mol 1mol HO HO O O STEP1 STEP STEP Grams Moles Moles Grams KNOWN g STEP 1: moles = GFM MolesKNOWN x STEP : = Coefficient Coefficient KNOWN STEP : g = moles GFM KNOWN UNKNOWN UNKNOWN UNKNOWN

4 TEST 4 REVIEW Page 4 Limiting Reactants and Percent Yield. The reactants that get used up first are limiting reactants. The limiting reactants can be used to predict the theoretical yield. The percent yield is determined by comparing the theoretical yield to the actual yield. See below. If 68.5 kg of CO reacts with 8.60 kg of H to produce 5.7 kg of CH OH, what is the percent yield? Step 1: Write a balanced equation. H (g) + CO(g)! CH OH(R) Step : Identify the limiting reactant by calculating the actual number of moles of each reactant. Then use either amount to determine the theoretical amount of the other 1000g 1mol CO 685. kg CO = mol CO 1kg 8. 01g CO actual 1000g 1mol H 860. kg H = mol H 1kg 016. g H mol H mol CO = mol H theoretical 1mol CO Since mol of H is more than we have, H is the limiting reactant Step : Use the limiting reactant to complete the calculation of the theoretical yield 1mol CHOH. 04g CHOH 1kg mol H = 68. 4kg CHOH mol H 1mol CH OH 1000g Step 4: Calculate the percent yield by comparing the actual to the theoretical 57. kg CHOH 100% = 5. % 68. 4kg CH OH Answer the questions below by circling the number of the correct response 1. The molecular mass of CO is the same as the molecular mass of (1) CO () C H 6 () SO (4) C H 8. Which is an empirical formula? (1) C H () Al Cl 6 () C H 4 (4) K O. A 60. gram sample of LiCl H O is heated in an open crucible until all of the water has been driven off. What is the total mass of LiCl remaining in the crucible? (1) 18 g () 4 g () 4 g (4) 60 g 4. Which is an empirical formula? (1) CH () C H 6 () C H 4 (4) C 4 H 8 5. A compound with a molecular mass of 4 contains hydrogen and oxygen in a ratio of 1:1. The molecular formula of the compound is (1) HO () OH () H O (4) HOH 6. The empirical formula of a compound is CH. Its molecular mass could be (1) 1 () 51 () 40 (4) What is the percentage by mass of bromine in CaBr? (1) 0% () 40% () 60% (4) 80% 8. The percent by mass of Li in LiNO (formula mass = 69) is closest to (1) 6% () 10% () 18% (4) 0% 9. The percent by mass of oxygen in CO is approximately (1) 7% () 57% () 4% (4) 17% 10. What is the empirical formula of the compound whose molecular formula is C 6 H 1 O 6? (1) C 1 H 4 O 1 () C H 4 O () C 6 H 1 O 6 (4) CH O 11. The percent by mass of aluminum in Al O is approximately (1) 18.9 () 5.4 () 47.1 (4) A compound contains nitrogen and oxygen in a ratio of 1:1. The molecular mass of the compound could be (1) 14 () 16 () 0 (4) The percent by mass of oxygen in Na SO 4 (formula mass = 14) is closest to (1) 11% () % () 45% (4) 64% 14. What is the ratio by mass of sulfur to oxygen in SO? (1) 1:1 () 1: () 1: (4) 1:4

5 TEST 4 REVIEW Page What is the mass in amu of 1.00 molecule of O gas? (1) 11. () 16.0 ().4 (4) What is the formula mass of CuSO 4 5H O? (1) 160. amu () 178 amu () 186 amu (4) 50. amu 17. What is the molecular formula of a compound whose empirical formula is CH 4 and molecular mass is 16? (1) CH 4 () C 4 H 8 () C H 4 (4) C 8 H The percent by mass of hydrogen in NH is equal to (1) () () 100 (4) The formula mass of NH 4 Cl is (1).4 amu () 5.5 amu () 8.0 amu (4) 95.5 amu 0. An example of an empirical formula is ( 1) C H, () H O, () C Cl, (4) CaCl 1. A compound has an empirical formula of CH and a molecular mass of 56. Its molecular formula is (1) C H 4, () C H 6, () C 4 H 8, (4) C 5 H 10.. What is the percent by mass of hydrogen in NH (formula mass = 17.0)? (1) 5.9% () 17.6% () 1.4% (4) 8.4%. The empirical formula of a compound is CH and its molecular mass is 70. What is the molecular formula of the compound? (1) C H () C H 4 () C 4 H 10 (4) C 5 H The percent by mass of nitrogen in Mg(CN) is equal to (1) 14 / , () 14 / , () 8 / , (4) 8 / What is the percent by mass of oxygen in Fe O (formula mass = 160)? (1) 16% () 0.% () 56% (4) 70.% 6. Which formulas could represent the empirical formula and the molecular formula of a given compound? (1) CH O, C 4 H 6 O 4 () CHO C 6 H 1 O 6 () CH 4, C H 8 (4) CH, C H 6 7. The percent by mass of carbon in CO is equal to (1) 44 / 1 100, () 1 / , () 8 / 1 100, (4) 1 / The empirical formula of a compound is CH 4. The molecular formula of the compound could be (1) CH 4, () C H 6, () C H 8, (4) C 4 H What is the percent by mass of oxygen in CH OH? (1) 50.0 () 44.4 ().0 (4) The approximate percent by mass of potassium in KHCO is (1) 19 %, () 4 %, () 9 %, (4) 61 % 1. A compound has an empirical formula of CH and a molecular mass of 56. What is its molecular formula? (1) CH () C H 6 () C H 4 (4) C 4 H 8. What is the percent by mass of hydrogen in CH COOH (formula mass = 60.)? (1) 1.7% () 6.7% () 5.0% (4) 7.1%. What is the percentage by mass of oxygen in CuO? (1) 16% () 5% () 0% (4) 50% 4. When the equation H + N! NH is completely balanced using smallest whole numbers, the sum of all the coefficients will be (1) 6 () 7 () (4) 1 5. A 10.0 gram sample of a hydrate was heated until all the water of hydration was driven off. The mass of anhydrous product remaining was 8.00 grams What is the percent of water in the hydrate? (1) 1.5% () 0.0% () 5.0% (4) 80.0% 6. A compound has the empirical formula NO. Its molecular formula could be (1) NO () N O () N 4 O (4) N 4 O 4 7. When the equation H + Fe O 4! Fe + H O is completely balanced using smallest whole numbers the coefficient of H would be (1) 1 () () (4) 4 8. When the equation C H 4 + O! CO + H O is correctly balanced, using smallest whole-numbered coefficients, the sum of all the coefficients is (1) 16 () 1 () 8 (4) 4 9. When the equation NH + O! HNO + H O is completely balanced using smallest whole numbers, the coefficient of O would be (1) 1 () () (4) When the equation Na(s) + H O(R)! NaOH(aq) + H (g) is correctly balanced using smallest whole numbers, the coefficient of the water is (1) 1 () () (4) Given the reaction: N (g) + H (g) W NH (g) What is the ratio of moles of H (g) consumed to moles of NH (g) produced? (1) 1: () : () : (4) 6:6 4. When the equation Al(s) + O (g)! Al O (s) is correctly balanced using the smallest whole numbers, the coefficient of Al(s) is (1) 1 () () (4) 4 4. Given the unbalanced equation: Al (SO 4 ) + Ca(OH)! Al(OH) + CaSO 4., when the equation is completely balanced using the smallest whole-number coefficients, the sum of the coefficients is (1) 15 () 9 () (4) Which quantity is equivalent to 9 grams of LiF? (1) 1.0 mole ().0 moles () 0.0 mole (4) 1.5 moles 45. What is the total number of molecules contained in 0.50 mole of O at STP [NOTE: 1 mol = particles]? (1) () ().0 10 (4) 1.5 x At STP, what mass of CH 4 has the same number of molecules as 64 grams of SO? (1) 16 g () g () 64 g (4) 18 g

6 TEST 4 REVIEW Page What is the total number of moles contained in 115 grams of C H 5 OH? (1) 1.00 () 1.50 ().00 (4) How many moles of water are contained in 0.50 mole of CuSO 4 5H O? (1) 1.5 () 40.0 () 4.50 (4) What is the mass of.0 10 atoms of neon [NOTE: 1 mol = particles]? (1) 1.0 g () 10. g () 0.50 g (4) 0. g 50. Which represents the greatest mass of chlorine (1) 1 mole of chlorine () 1 atom of chlorine () 1 gram of chlorine (4) 1 molecule of chlorine 51. What is the total mass of iron in 1.0 mole of Fe O? (1) 160 g () 7 g () 11 g (4) 56 g 5. What is the mass, in grams, of 1.0 mole of (NH 4 ) S? (1) 50. () 54 () 64 (4) What is the gram atomic mass of the element chlorine? (1) 17 g () 5 g () 5 g (4) 70. g 54. The mass in grams of 1.00 mole of CaSO 4 H O is (1) 17 g () 154 g () 16 g (4) 118 g 55. Which compound contains the greatest percentage of oxygen by mass? (1) BaO () MgO () CaO (4) SrO 56. The precent by mass of oxygen in MgO (formula mass = 40) is closest to (1) 16% () 40% () 4% (4) 60% 57. The symbol (aq) after a chemical formula means (1) solid or precipitate, () liquid, () gas, (4) aqueous or dissolved. 58. In the reaction, AgNO + NaCl! AgCl + NaNO, the reactants are (1) AgCl and NaNO, () AgNO and NaCl, () Ag and Na, (4) Cl and NO 6. Given the reaction: 4Al + O! Al O How many moles of Al O will be formed when 7 grams of Al reacts completely with O? (1) 1.0 ().0 () 0.50 (4) A compound consists of 85% silver and 15% fluorine by mass. What is its empirical formula? (1) AgF () Ag F () AgF (4) Ag 6 F 64. Given the reaction: Cu + 4HNO! Cu(NO ) + H O + NO What is the total mass of H O produced when grams of Cu is completely consumed? (1) 9.0 g () 18 g () 6 g (4) 7 g For each of the reactions described in questions 65-71, write the correct number to indicate whether the reaction type is (1) DECOMPOSITION, () DIRECT COMBINATION, () SINGLE REPLACEMENT, or (4) DOUBLE REPLACEMENT 65. A reaction occurs in which only one reactant is present. 66. A metal reacts with an acid. (Fe + 6HCl! FeCl + H ) 67. Magnesium burns. 68. Two salt solutions react with each other. 69. Two elements unite to form a compound. 70. A compound breaks down. 71. HCl + NaOH! NaCl + H O 7. If 00.0 kg of iron ore are reacted with 00.0 kg of carbon monoxide to produce kg of pure iron,[fe O (s) + CO(g) ÿ Fe(s) + CO (g)] what is the percent yield? (1) % ().8 % () % (4) 95.8 % Answer questions by referring to the equation below: KClO (s) MnO ( s) KCl(s) + O (g) 59. The symbol under the yield sign indicates that (1) the reaction is exothermic, () the reaction is endothermic, () a solid precipitate forms, (4) heat is a product of the reaction. 60. MnO (s) is written above the yield sign because MnO (s) is (1) a reactant, () a product, () neither a reactant nor a product, (4) both a reactant and a product Answers Given the balanced equation: NaOH + HCl! NaCl + H O, what is the total number of grams of H O produced when 116 grams of the product, NaCl, is formed? (1) 9.0 g () 18 g () 6 g (4) 54 g

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