1 T-6 Tutorial 2 FORMULAS, PERCENTAGE COMPOSITION, AND THE MOLE FORMULAS: A chemical formula shows the elemental composition of a substance: the chemical symbols show what elements are present and the numerical subscripts show how many atoms of each element there are in a formula unit. Examples: NaCl: one sodium atom, one chlorine atom in a formula unit CaCl 2 : one calcium atom, two chlorine atoms in a formula unit Mg 3 N 2 : three magnesium atoms, two nitrogen atoms in a formula unit The presence of a metal in a chemical formula indicates an ionic compound, which is composed of positive ions (cations) and negative ions (anions). A formula with only nonmetals indicates a molecular compound (unless it is an ammonium, NH 4 +, compound). Only ionic compounds are considered in this Tutorial. There are tables of common ions in your lecture text, p 56 (cations) and p 57 (anions). A combined table of these same ions can be found on the inside back cover of the lecture text. A similar list is on the next page; all formulas needed in this and subsequent Tutorial problems can be written with ions from this list. Writing formulas for ionic compounds is very straightforward: TOTAL POSITIVE CHARGES MUST BE THE SAME AS TOTAL NEGATIVE CHARGES. The formula must be neutral. The positive ion is written first in the formula and the name of the compound is the two ion names. EXAMPLE: Write the formula for potassium chloride. The name tells you there are potassium, K +, and chloride, Cl, ions. Each potassium ion is +1 and each chloride ion is -1: one of each is needed, and the formula for potassium chloride is KCl. "1" is never written as a subscript. EXAMPLE: Write the formula for magnesium hydroxide. This contains magnesium, Mg 2+, and hydroxide, OH, ions. Each magnesium ion is +2 and each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium hydroxide is Mg(OH) 2. The (OH) 2 indicates there are two OH ions. In a formula unit of Mg(OH) 2, there are one magnesium ion and two hydroxide ions; or one magnesium, two oxygen, and two hydrogen atoms. The subscript multiplies everything in ( ). EXAMPLE: Write the formula for aluminum sulfate. This contains aluminum, Al 3+, and sulfate, SO 4 2, ions. The lowest common multiple of 3 and 2 is 6, so we will need six positive and six negative charges: two Al 3+ and three SO 4 2 ions, and the formula for aluminum sulfate is Al 2 (SO 4 ) 3. Then, in a formula unit of Al 2 (SO 4 ) 3 there are two aluminum ions and three sulfate ions; or two aluminum, three sulfur, and twelve oxygen atoms.
2 COMMON POSITIVE IONS COMMON NEGATIVE IONS COMMON NEGATIVE IONS H + hydrogen [Fe(CN) 6 ] 3 ferricyanide C 2 H 3 O 2 acetate NH 4 + ammonium [Fe(CN) 6 ] 4 ferrocyanide CN cyanide Li + lithium 3 PO 4 phosphate CNO cyanate Na + sodium 2 HPO 4 hydrogen phosphate SCN thiocyanate K + potassium H 2 PO 4 dihydrogen phosphate ClO hypochlorite Mg 2+ magnesium 2 CO 3 carbonate ClO 3 chlorate Ca 2+ calcium HCO 3 hydrogen carbonate ClO 4 perchlorate Sr 2+ strontium 2 SO 3 sulfite IO 3 iodate Ba 2+ barium HSO 3 hydrogen sulfite MnO 4 permanganate Al 3+ aluminum 2 SO 4 sulfate NO 2 nitrite Sn 2+ tin(ii) HSO 4 hydrogen sulfate NO 3 nitrate Sn 4+ tin(iv) 2 S 2 O 3 thiosulfate OH hydroxide Pb 2+ lead(ii) 1 2 CrO 4 chromate IO 4 periodate Bi 3+ bismuth 2 Cr 2 O 7 dichromate H hydride Cr 3+ chromium(iii) 2 O 2 oxide F fluoride Mn 2+ manganese(ii) 3 2 O 2 peroxide Cl chloride Fe 2+ iron(ii) S 2 sulfide Br bromide Fe 3+ iron(iii) HS hydrogen sulfide I iodide Co 2+ cobalt(ii) 4 Ni 2+ nickel(ii) 5 Cu + copper(i) Cu 2+ copper(ii) 1 There is also a lead(iv) Ag + silver 2 There is also a chromium(ii) Zn 2+ Cd Hg 2 2+ Hg 2 zinc 3 There is also a manganese(iii) cadmium 4 There is also a cobalt(iii) mercury(i) 5 There is also a nickel(iii) mercury(ii)
3 T-8 PERCENTAGE COMPOSITION: Imagine a class of 40 boys and 60 girls: 100 students total. 40/100 of the students are boys and 60/100 girls; or, 40% boys and 60% girls. The "%" sign means percent, or parts per 100. Suppose a class has 10 boys and 15 girls, for a total of 25 students. If you want to find out how many boys there would be per 100 students, keeping the same ratio of boys to girls, the following proportion can be set up: 10 boys X boys = 25 students 100 students This means: "if there are 10 boys per 25 students, then there would be X boys per 100 students." Solving this equation, 10 boys X = x 100 students = 40 boys 25 students There would be 40 boys per 100 students, or 40% boys. Percent calculation is based on this kind of proportion. However, you do not have to set up the proportion each time you want to find percent composition. Simply divide the number of parts you are interested in (number of boys) by the total number of parts (number of students), and multiply by 100. This is exactly what we did to find X in the above example. DIVIDE THE PART BY THE WHOLE AND MULTIPLY BY 100 Suppose a mixture contains 12.4 g salt, 15.3 g sugar and 50.3 g sand, for a total 78.0 g of mixture. The percentage of each component in the mixture can be found by dividing the amount of each (PART) by the total amount (WHOLE) and multiplying by 100: 12.4 g 15.3 g x 100 = 15.9% salt x 100 = 19.6% sugar 78.0 g 78.0 g 50.3 g x 100 = 64.5% 78.0 g The sum of these percentages is 100%. Everything must total 100%.
4 T-9 Percentage is written without units but the units are understood to be "parts of whatever per 100 total parts." The "parts" and "whole" may be in any units as long as they are both in the same units. In the case of the above mixture, units could be put on the numbers and the percentages then used as conversion factors. For example, the 15.9% salt could become 15.9 g salt pounds mixture 15.9 tons salt or or g mixture 15.9 pounds salt tons mixture Et cetera. We do not usually care about composition of salt/sugar/sand mixtures, but we do care about percentage composition of pure chemical substances. Unless otherwise stated we always mean percent by mass (weight). The formula mass of a substance is the sum of the atomic masses of all the atoms in the formula. For example, the formula mass of sodium chloride, NaCl, 58.44, is the sum of the atomic mass of sodium, 22.99, and the atomic mass of chlorine, The formula mass of sodium sulfate, Na 2 SO 4, is found as follows: 2 Na: 2 x = g Na or lbs Na 1 S: 1 x = g S or lbs S 4 O: 4 x = g O or lbs O Formula mass = g Na 2 SO lbs Na 2 SO 4 Atomic mass and formula mass are in atomic mass units, but are usually written without units. However, since these are relative masses of the atoms any mass unit can be used, as shown by the two columns to the right of the arrows, above. In terms of atomic masses: in a total of parts by mass of sodium sulfate, parts are sodium, parts are sulfur, and parts are oxygen. In terms of a gram mass unit (molar mass): in a total of g of sodium sulfate, g are sodium, g are sulfur, and g are oxygen. In terms of pound mass unit: in a total of lbs of sodium sulfate, lbs are sodium, lbs are sulfur, and lbs are oxygen.
5 T-10 Percent composition can be determined with the mass ratios from the formula mass: Sodium: x 100 = 32.37% Na g Na or lbs Na Sulfur: x 100 = 22.57% S g S or lbs S Oxygen: x 100 = 45.06% O g O or lbs O Total = % g total lbs total As mentioned before, percentage is written without any units but the units are understood to be "parts of whatever per 100 total parts". Since these are relative masses any mass unit can be used, as shown by the two columns to the right of the arrows, above. In terms of percentage: in a total of parts by mass of sodium sulfate, parts are sodium, parts are sulfur, and parts are oxygen. In terms of a gram mass unit (molar mass): in a total of g of sodium sulfate, g are sodium, g are sulfur, and g are oxygen. In terms of a pound mass unit: in a total of lbs of sodium sulfate, lbs are sodium, lbs are sulfur, and lbs are oxygen. In the preceding calculations, the percent of each element in the substance was found. The percent of groups of elements may also be calculated. Suppose you want percent sulfate, SO 4 2, in sodium sulfate. The formula mass of sodium sulfate is and of that total = is sulfate. Thus, Sulfate: x 100 = 67.63% SO Notice: this is the sum of the percents sulfur and oxygen: 22.57% S % 0 = 67.63% SO 4 2. We may say sodium sulfate is 32.37% sodium, 22.57% sulfur, and 45.06% oxygen; or, 32.37% sodium and 67.63% sulfate. The total is 100% in either case.
6 T-11 We now have two sets of numbers which show mass relationships among the elements in sodium sulfate: (1) the formula mass and (2) the percentage composition. From the formula mass on page T-9 various mass ratios, that is, conversion factors, may be obtained: g Na lbs Na 2 SO g S or or and others g Na 2 SO lbs S g Na From the percentages on page T-10 various mass ratios may also be obtained: g Na lbs Na 2 SO g S or or and others g Na 2 SO lbs S g Na We can now solve problems like the following: EXAMPLE: What mass of sulfur is contained in 17.2 g of sodium sulfate? Use formula mass ratios on page T-9 for the appropriate conversion factor: g S 17.2 g Na 2 SO 4 x = 3.88 g S g Na 2 SO 4 OR, use percentage composition ratios on page T-10: g S 17.2 g Na 2 SO 4 x = 3.88 g S g Na 2 SO 4
7 T-12 EXAMPLE: What mass of sodium is combined with 10.0 g of sulfur in sodium sulfate? You may use a conversion factor from the formula masses OR percentage composition: OR g Na 10.0 g S x = 14.3 g Na g S g Na 10.0 g S x = 14.3 g Na g S Obviously, you do not have to calculate percent composition every time you do a problem like this: use formula mass ratios; if the percentage composition is given, then use it. THE MOLE: The most important concept you encounter this semester in Chemistry is the mole. The mole may be defined: The mole is the amount (mass) of a substance which contains the same number of units (atoms, molecules, ions) as there are atoms in exactly 12 mass units of the carbon-12, 12 C, isotope. It does not matter what mass units are used as long as the same mass units are used for the substance and the carbon-12. This is a strict definition of the mole, but it is operationally useless. A mole of a monoatomic element is the atomic mass taken with mass units; a mole of anything else is the formula mass taken with mass units. The kind of mole you get depends on the mass units you use. For example: grams Na = 1 gram-mole Na has the same number of Na atoms as the number of atoms in 12 grams of 12 C milligrams Cu = 1 milligram-mole Cu has the same number of Cu atoms as the number of atoms in 12 milligrams of 12 C grams Cl 2 = 1 gram-mole Cl 2 has the same number of Cl 2 molecules as the number of atoms in 12 grams of 12 C lbs I 2 = 1 lb-mole I 2 has the same number of I 2 molecules as the number of atoms in 12 lbs of 12 C grams NaCl = 1 gram-mole NaCl has the same number of NaCl formula units as the number of atoms in 12 grams of 12 C.
8 T tons Na 2 SO 4 = 1 ton-mole Na 2 SO 4 has the same number of Na 2 SO 4 formula units as the number of atoms in 12 tons of 12 C g H 2 O = 1 gram-mole H 2 O has the same number of H 2 O molecules as the number of atoms in 12 g of 12 C kg PCl 3 = 1 kilogram-mole PCl 3 has the same number of PCl 3 molecules as the number of atoms in 12 kg of 12 C. Chemists generally use the gram-mole, written mole, abbreviated mol. This is the mole your textbook uses and is the only one for which Avogadro's Number is applicable. In one grammole of a substance there is Avogadro's Number of units: x of them. Unless further qualification is noted the term "mole" is understood to mean the gram-mole. Each of the equalities above becomes a conversion factor: And so on g Na 1 mol Na g Na = 1 mol Na becomes or 1 mol Na g Na EXAMPLE: How many moles is g of sodium metal? 1 mol Na g Na x = mol Na g Na EXAMPLE: What is the mass of mol of sodium sulfate? 142 g Na 2 SO mol Na 2 SO 4 x = 51.4 g Na 2 SO 4 1 mol Na 2 SO 4 The number of significant figures used in the atomic mass or formula mass should be at least as many as in the given (measured) data. EXAMPLE: How many atoms of copper are in a 125 mg sample of copper metal? When a problem asks "how many atoms or molecules" or "what is the mass of an atom or molecule," you must use Avogadro's Number. The units associated with that Number then depend on the problem. In this problem you are looking for atoms of copper so the units will be:
9 T x atoms Cu 1 mol Cu The number of atoms of copper is related to moles of copper; moles of copper is related to grams of copper; the mass of copper given in the problem is in milligrams. The conversion sequence will be: mg Cu g Cu mol Cu atoms Cu 1 g Cu 1 mol Cu x atoms Cu 125 mg Cu x x x = 1000 mg Cu 63.5 g Cu 1 mol Cu 1.19 x atoms Cu EXAMPLE: What is the mass of one water molecule? This time the units associated with Avogadro's Number are x H 2 O molecules 1 mol H 2 O We are converting from a number of molecules (1, an exact number) to mass: 1 mol H 2 O 18.0 g H 2 O 1 H 2 O molecule x x = x H 2 O molecules 1 mol H 2 O 2.99 x g H 2 O If four significant figures had been used for the formula mass of water, 18.02, the answer would have been x g H 2 O. In this case the precision of the answer depends on the conversion factors, since the given data is an exact number. Chemical formulas should be considered in terms of moles of atoms combined: the formula NaCl means there is one mole of sodium combined with one mole of chlorine; the formula H 2 O means there are two moles of hydrogen combined with one mole of oxygen to form one mole of water.
10 T-15 The formula mass for sodium sulfate was calculated on page T-9. Referring to those numbers, the formula Na 2 SO 4 means: two moles sodium (45.98 g), one mole sulfur (32.06 g), and four moles oxygen (64.00 g) combine to form one mole of sodium sulfate ( g). Let us use this concept and do the first problem on page T-11 a different way: EXAMPLE: What mass of sulfur is contained in 17.2 g of sodium sulfate? 1 mol Na 2 SO 4 1 mol S 32.1 g S 17.2 g Na 2 SO 4 x x x = 3.89 g S 142 g Na 2 SO 4 1 mol Na 2 SO 4 1 mol S Please notice: The same numbers are used in the same places, so the overall arithmetic is the same. The thought process is different: on page T-11 we thought in terms of mass ratios, now we are thinking in terms of mole ratios. This answer is slightly different from the one obtained on page T-11. Can you see why? EMPIRICAL (SIMPLEST) FORMULA: The empirical formula is the formula expressed as the lowest whole number ratio of atoms; it may or may not be the same as the formula. The empirical formula and formula of water, H 2 O, are the same; the empirical formula of phosphorous pentoxide is P 2 O 5, but the formula is P 4 O 10. The empirical formula is derived solely from experimental, percentage composition data. On page T-10 we calculated percentage composition from a formula; let us now calculate an empirical formula from percentage composition data. Fundamental to this calculation is the fact: chemical formulas show mole ratios of elements. EXAMPLE: Calculate the empirical formula for a compound with the analysis: 32.37% Na 22.57% S 45.06% O Remember percentage means "parts of whatever per 100 total parts," so let us translate these percentages into grams of each element out of a total 100 grams of compound. Then convert these masses to moles: 1 mol Na g Na x = mol Na g Na 1 mol S g S x = mol S g S 1 mol O g O x = mol O g O
11 T-16 Please note: these are the number of moles of each element contained in g of compound. Now find the ratio of moles of each element to the smallest number of moles, by dividing each number of moles by the smallest: mol Na = mol Na / mol S mol S mol S = mol S mol O = mol O / mol S mol S These ratios say there are 2 mol Na per 1 mol S; and 4 mol O per 1 mol S. The lowest wholenumber ratio of moles is: Na 2 S 1 O 4 ; and since we do not write "1" as a subscript, the formula is: Na 2 SO 4. Sometimes these two steps do not give whole number ratios. In that case an additional step is needed, as shown below. EXAMPLE: Calculate the empirical formula for a compound with the analysis: 29.1% Na 40.5% S 30.4% O Let us do as with the previous EXAMPLE: (1) translate the percents into grams of the element per 100 g of compound and find moles of each element; then (2) find the ratio of moles of each element to the smallest number of moles: 1 mol Na 1.27 mol Na 29.1 g Na x = 1.27 mol Na = 1.01 mol Na/mol S 23.0 g Na 1.26 mol S 1 mol S 1.26 mol S 40.5 g S x = 1.26 mol S = g S 1.26 mol S 1 mol O 1.90 mol O 30.4 g O x = 1.90 mol O = 1.51 mol O/mol S 16.0 g O 1.26 mol S
12 T-17 These ratios say there is 1 mol Na per 1 mol S, and 1.5 mol O per 1 mol S, which are not whole number ratios. To make them whole number ratios we must multiply each by 2, giving ratios of 2 mol Na per 2 mol S and 3 mol O per 2 mol S; and the formula is Na 2 S 2 O 3. Thus, in cases where steps (1) and (2) do not lead to whole numbers, (3) multiply each ratio by some integer (2, 3, etc.) to make all the ratios whole numbers. 1) Write formulas for the following: a) bismuth nitrate j) zinc phosphate s) copper(ii) oxide b) calcium hydride k) magnesium carbonate t) lithium nitrate c) mercury(ii) cyanide l) tin(ii) chloride u) aluminum sulfide d) strontium iodide m) lead(ii) bromide v) iron(ii) phosphate e) tin(iv) hydroxide n) ammonium dichromate w) barium sulfate f) potassium sulfite o) nickel(ii) chromate x) silver nitrate g) sodium carbonate p) cadmium cyanate y) copper(i) sulfide h) cobalt(ii) nitrite q) mercury(i) chloride z) iron(iii) sulfate i) chromium(iii) iodide r) manganese(ii) periodate 2) Calculate the percentage composition, by mass, of the following. Use four significant figures. a) potassium phosphate e) zinc chloride b) ammonium cyanide f) aluminum nitrate c) sodium hydrogen sulfate g) chromium(iii) hydroxide d) potassium perchlorate h) copper(ii) carbonate 3) Using the percentages from problem (2) calculate the following: a) grams of phosphate in 65.5 g of potassium phosphate b) milligrams of nitrogen in mg of ammonium cyanide c) pounds of sodium in 10.0 lbs of sodium hydrogen sulfate d) grams of chlorine in 37.5 g of potassium perchlorate e) tons of zinc in tons of zinc chloride f) tons of nitrate in 125 tons of aluminum nitrate g) milligrams of oxygen in g chromium(iii) hydroxide h) pounds carbon in 138 lbs copper(ii) carbonate
13 T-18 4) Convert each of the following to moles: a) g of iron(iii) sulfate b) 3.67 x 10 2 g of lead(ii) nitrate c) x 10 3 g of tin(ii) bromide d) kg of ammonium chromate e) 175 g of manganese(ii) carbonate f) 8.5 x 10 6 g of zinc oxide g) mg of barium sulfate h) 3.55 x 10 5 mg of aluminum chloride 5) Calculate the mass (in grams) of each of the following: a) 1.00 mol of barium sulfate b) mol of zinc nitrate c) mol of sodium hydrogen carbonate d) mol of potassium permanganate e) 1.67 x 10 2 mol of sodium hydroxide f) x 10 4 mol of copper(i) chloride g) mol of magnesium chloride h) 1.45 x 10 2 mol of calcium bromide 6) Calculate the following: a) grams of manganese in 1.55 g of manganese(ii) nitrate b) tons of phosphorous in 5.73 tons of potassium phosphate c) moles of sulfur in 7.33 g of aluminum sulfate d) pounds of iron in 12.6 lbs of iron(ii) phosphate e) moles of nitrogen in 5.55 g of lead(ii) nitrate f) grams of nitrogen in 3.95 g of ammonium phosphate g) grams of iron in 1.37 kg of iron(iii) sulfide h) moles of oxygen in g of iron(iii) sulfate 7) Calculate the following: a) number of sodium atoms in 6.79 g of sodium metal b) number of sodium ions in 7.77 g of sodium phosphate c) number of water molecules in 15.5 mg of water d) number of cobalt atoms in mg of cobalt metal e) number of ammonia molecules, NH 3, in 375 kg of ammonia f) number of gold atoms in 1.00 x 10 6 mg of gold g) number of ozone molecules, O 3, in 1.73 x 10 8 g of ozone h) mass, in grams, of ten water molecules i) mass, in grams, of one mercury atom j) mass, in milligrams, of three sulfuric acid, H 2 SO 4, molecules k) mass, in kilograms, of 1.25 x sulfate ions l) mass, in kilograms, of 2.57 x ammonium ions
14 T-19 8) Calculate the empirical formulas for the compounds with the following analyses: a) 34.4% Fe 65.6% Cl b) 29.4% Ca 23.6% S 47.0% O c) 28.7% K 1.5% H 22.8% P 47.0% O d) 28.0% Fe 24.1% S 48.0% O e) 88.8% Cu 11.2% O f) 72.3% Fe 27.7% O g) 69.9% Fe 30.1% O h) 11.1% N 3.2% H 41.3% Cr 44.4% O i) 68.4% Cr 31.6% O Answers to Problems 1) a) Bi(NO 3 ) 3 h) Co(NO 2 ) 2 o) NiCrO 4 u) Al 2 S 3 b) CaH 2 i) CrI 3 p) Cd(CNO) 2 v) Fe 3 (PO 4 ) 2 c) Hg(CN) 2 j) Zn 3 (PO 4 ) 2 q) Hg 2 Cl 2 w) BaSO 4 d) SrI 2 k) MgCO 3 r) Mn(IO 4 ) 2 x) AgNO 3 e) Sn(OH) 4 l) SnCl 2 s) CuO y) Cu 2 S f) K 2 SO 3 m) PbBr 2 t) LiNO 3 z) Fe 2 (SO 4 ) 3 g) Na 2 CO 3 n) (NH 4 ) 2 Cr 2 O 7 2) a) 55.26% K; 14.59% P; 30.15% O for K 3 PO 4 b) 63.59% N; 9.152% H; 27.26% C for NH 4 CN c) 19.15% Na; % H; 26.71% S; 53.30% O for NaHSO 4 d) 28.22% K; 25.59% Cl; 46.19% O for KCl0 4 e) 47.98% Zn; 52.02% Cl for ZnCl 2 f) 12.67% Al; 19.73% N; 67.60% O for Al(NO 3 ) 3 g) 50.47% Cr; 46.59% O; 2.935% H for Cr(OH) 3 h) 51.43% Cu; 9.720% C; 38.85% O for CuCO 3 3) a) g PO 4 e) tons Zn b) mg N f) tons NO 3 c) 1.92 lbs Na g) mg O d) 9.60 g Cl h) 13.4 lbs C
15 T-20 4) a) mol Fe 2 (SO 4 ) 3 e) 1.52 mol MnCO 3 b) 1.11 x 10 4 mol Pb(NO 3 ) 2 f) 1.0 x 10 7 mol ZnO c) mol SnBr 2 g) x 10 6 mol BaSO 4 d) mol (NH 4 ) 2 CrO 4 h) 2.66 mol AlCl 3 5) a) 233 g BaSO 4 e) g NaOH b) g Zn(NO 3 ) 2 f) g CuCl c) 13.0 g NaHCO 3 g) g MgCl 2 d) g KMnO 4 h) 2.90 x 10 4 g CaBr 2 6) a) g Mn in Mn(NO 3 ) 2 e) mol N in Pb(NO 3 ) 2 b) ton P in K 3 PO 4 f) 1.11 g N in (NH 4 ) 3 PO 4 c) mol S in Al 2 (SO 4 ) 3 g) 735 g Fe in Fe 2 S 3 d) 5.90 lbs Fe in Fe 3 (PO 4 ) 2 h) mol O in Fe 2 (SO 4 ) 3 7) a) 1.78 x Na atoms g) 2.17 x O 3 molecules b) 8.56 x Na 1+ ions in Na 3 PO 4 h) x g H 2 O c) 5.19 x H 2 O molecules i) x g Hg d) x Co atoms j) x mg H 2 SO 4 e) 1.33 x NH 3 molecules k) 1.99 x kg SO 4 f) 3.06 x Au atoms l) kg NH 4 1 8) a) FeCl 3 b) CaSO 4 c) KH 2 PO 4 d) Fe 2 (SO 4 ) 3 e) Cu 2 O f) Fe 3 O 4 g) Fe 2 O 3 h) (NH 4 ) 2 Cr 2 O 7 i) Cr 2 O 3
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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
1 POLYATOMIC IONS We have seen that atoms can lose or gain electrons to become ions. Groups of atoms can also become ions. These groups of atoms are called polyatomic ions. Examples: O hydroxide ion NO
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
Chapter 4 Compounds and Their Bonds 4.1 Octet Rule and Ions Octet Rule An octet is 8 valence electrons. is associated with the stability of the noble gases. He is stable with 2 valence electrons (duet).
6 CEMICAL NAMES AND FORMULAS SECTION 6.1 INTRODUCTION TO CEMICAL BONDING (pages 133 137) This section explains how to distinguish between ionic and molecular compounds. It also defines cation and anion
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
Chem 1100 Chapter Three Study Guide Answers Outline I. Molar Mass and Moles A. Calculations of Molar Masses B. Calculations of moles C. Calculations of number of atoms from moles/molar masses 1. Avagadro
Experiment 5 Chemical Reactions OBJECTIVES 1. To observe the various criteria that are used to indicate that a chemical reaction has occurred. 2. To convert word equations into balanced inorganic chemical
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
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
Nomenclature and the Periodic Table To name compounds and to determine molecular formulae from names a knowledge of the periodic table is helpful. Atomic Number = number of protons Mass Number = number
Instructions for Conversion Problems For every conversion problem Write the number in the problem down with unit and a multiplication sign Decide which conversion factor you should use, Avagadro s or molar
Chapter 9 Practice Test - Naming and Writing Chemical Formulas Matching Match each itme with the correct statement below. Match each item with the correct statement below. a. monatomic ion f. cation b.
207 Oxidation- reduction (redox) reactions Chapter 12: Oxidation and Reduction. At different times, oxidation and reduction (redox) have had different, but complimentary, definitions. Compare the following
Name Writing and Balancing Chemical Equations Period When a substance undergoes a chemical reaction, chemical bonds are broken and new bonds are formed. This results in one or more new substances, often
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
Polyatomic Ions Worksheet Polyatomic Practice 1. Name or write the formula for the following polyatomic ions sulfate - CO nitrite MnO perphosphate - SO hypoiodite BrO chlorite - CO phosphite - PO percarbonate
CHAPTER 9 Chemical Names and Formulas 9.1 Naming Ions Monatomic Ions: a single atom with a positive or negative charge Cation (rules): listed first Anion (rules): ide ending Transition Metals have a varying
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
Chapter 3 Stoichiometry 3-1 Chapter 3 Stoichiometry In This Chapter As you have learned in previous chapters, much of chemistry involves using macroscopic measurements to deduce what happens between atoms
Molecules, Compounds, and Chemical Equations (Chapter 3) Chemical Compounds 1. Classification of Elements and Compounds Types of Pure Substances (Figure 3.4) Elements -- made up of only one type of atom
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
Chapter 2: Chemical Compounds and Bonding Section 2.1: Ionic Compounds, pages 22 23 1. An ionic compound combines a metal and a non-metal joined together by an ionic bond. 2. An electrostatic force holds
CHEMICAL NOMENCLATURE Chemical nomenclature The process of giving unambiguous chemical formulas or chemical names to elements and compounds Introduction Chemistry is the study of matter (elements and compounds)
Stoichiometry Review There are 20 problems in this review set. Answers, including problem set-up, can be found in the second half of this document. 1. N 2 (g) + 3H 2 (g) --------> 2NH 3 (g) a. nitrogen
Oxidation-Reduction Reactions Chapter 11 Electrochemistry Oxidation and Reduction Reactions An oxidation and reduction reaction occurs in both aqueous solutions and in reactions where substances are burned
T-27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient
How much does a single atom weigh? Different elements weigh different amounts related to what makes them unique. What units do we use to define the weight of an atom? amu units of atomic weight. (atomic
Name: Block: Date: Test Review: Chapter 8 Ionic Bonding Part 1: Fill-in-the-blank. Choose the word from the word bank below. Each word may be used only 1 time. electron dot structure metallic electronegativity
APPENDIX A Molarity of Ions in Solution ften it is necessary to calculate not only the concentration (in molarity) of a compound in aqueous solution but also the concentration of each ion in aqueous solution.
Aqueous Solutions and Solution Stoichiometry Water is the dissolving medium, or solvent. Some Properties of Water Water is bent or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration
Chem 115 POGIL Worksheet - Week 4 Moles & Stoichiometry Why? Chemists are concerned with mass relationships in chemical reactions, usually run on a macroscopic scale (grams, kilograms, etc.). To deal with
EPERIMENT 8: Activity Series (Single Displacement Reactions) PURPOSE a) Reactions of metals with acids and salt solutions b) Determine the activity of metals c) Write a balanced molecular equation, complete
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
CHEMICAL FORMULAS AND FORMULA WEIGHT CALCULATIONS 1. THE MEANING OF A CHEMICAL FORMULA A chemical formula is a shorthand method of representing the elements in a compound. The formula shows the formulas
Chapter 8: Chemical Equations and Reactions I. Describing Chemical Reactions A. A chemical reaction is the process by which one or more substances are changed into one or more different substances. A chemical
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
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)
Chapter 7 Page 1 Chapter 7: Chemical Reactions A chemical reaction: a process in which at least one new substance is formed as the result of a chemical change. A + B C + D Reactants Products Evidence that
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements