Chemical Formulas and Composition Stoichiometry

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2 Chemical Formulas and Composition Stoichiometry 2-1 (a) Stoichiometry is the description of the quantitative relationships among

(b) Composition stoichiometry describes the quantitative relationships among

Reaction stoichiometry describes the quantitative relationships among substances as they undergo chemical changes.

) 2-3 The common ions for each formula unit is listed below: (a) MgCl 2 contains Mg 2+ and Cl - ions (b) (NH 4 ) 2 CO 3 contains NH + 4

(space-filling; ball-and-stick) Both are composed of hydrogen, carbon, and oygen. Both have an oygen and hydrogen on the end.

2-7 Organic compounds are those that contain carbon-to-carbon bonds, carbon-to-hydrogen bonds, or both.

1 2 Chemical Formulas and Composition Stoichiometry 2-1 (a) Stoichiometry is the description of the quantitative relationships among elements in a compound and among substances as they undergo chemical change. (b) Composition stoichiometry describes the quantitative relationships among elements in compounds, e.g., in water, H 2 O, there are 2 hydrogen atoms for every 1 atom of oygen. Reaction stoichiometry describes the quantitative relationships among substances as they undergo chemical changes. (Reaction stoichiometry will be discussed in Chapter 3.) 2-3 The common ions for each formula unit is listed below: (a) MgCl 2 contains Mg 2+ and Cl - ions (b) (NH 4 ) 2 CO 3 contains NH + 4 and CO 2-3 ions (c) Zn(NO 3 ) 2 contains Zn 2+ and NO - 3 ions 2-5 Ethanol -CH 3 CH 2 OH Methanol-CH 3 OH (space-filling; ball-and-stick) (space-filling; ball-and-stick) Both are composed of hydrogen, carbon, and oygen. Both have an oygen and hydrogen on the end. The ethanol molecule has an additional carbon and two hydrogens. 2-7 Organic compounds are those that contain carbon-to-carbon bonds, carbon-to-hydrogen bonds, or both. Organic formulas given in Table 2-1 include: acetic acid- CH 3 COOH, methane- CH 4, ethane- C 2 H 6, propane- C 3 H 8, butane- C 4 H 10, pentane- C 5 H 12, benzene- C 6 H 6, methanol- CH 3 OH, ethanol- CH 3 CH 2 OH, acetone- CH 3 COCH 3, diethyl ether- CH 3 CH 2 COCH 2 CH Compounds from Table 2-1 that contain only carbon and hydrogen and are not shown in Figure 1-5: Compound Ball and stick model Compound Ball and stick model acetic acid- CH 3 COOH acetone- CH 3 COCH 3 methanol- CH 3 OH diethyl ether- CH 3 CH 2 COCH 2 CH 3 2-1

2 2-11 (a) Formula weight is the mass in atomic mass units of the simplest formula of an ionic compound and is found by adding the atomic weights of the atoms specified in the formula. The numerical amount for the formula weight is the equal to the numerical amount for the mass in grams of one mole of the substance. (b) Molecular weight is the mass in atomic mass units of one molecule of a substance that is molecular, rather than ionic. It is found by adding the atomic weights of the atoms specified in the formula. The numerical amount for the molecular weight is the equal to the numerical amount for the mass in grams of one mole of the substance. (c) Structural formula is the representation that shows how atoms are connected in a compound. (d) An ion is an atom or group of atoms that carries an electrical charge, which is caused by unequal numbers of protons and electrons. A postive ion is a cation. A negative ion is an anion The formulas for (a) through (d) are given in Table 2-1. (a) C 4 H 10 (b) CH 3 CH 2 OH (c) SO 3 (d) CH 3 COCH 3 (e) CCl We can find most of the names of the appropriate ions in Table 2-2. (a) magnesium chloride (b) iron(ii) nitrate (c) sodium sulfate (d) calcium hydroide (e) iron(ii) sulfate 2-17 Formulas are written to show the ions in the smallest ratio that gives no net charge. Compounds are electrically neutral. (a) NaOH, sodium hydroide (b) Al 2 (CO 3 ) 3, aluminum carbonate (c) Na 3 PO 4, sodium phosphate (d) Ca(NO 3 ) 2, calcium nitrate (e) FeCO 3, iron(ii) carbonate 2-19 (a) This chemical formula is incorrect. The atomic symbol for a potassium ion is K +, not P +. The correct chemical formula for potassium iodide is KI. (b) This chemical formula is correct. (c) The chemical formula is incorrect. The symbol for a silver ion is Ag +. The correct chemical formula for the carbonate ion is CO 3 2. Therefore, the chemical formula for silver carbonate is Ag 2 CO (a) Al(OH) 3 (b) MgCO 3 (c) ZnCO 3 (d) (NH 4 ) 2 SO 4 (e) ZnSO (a) CaCO 3 (b) Mg(OH) 2 (c) CH 3 COOH (d) NaOH (e) ZnO 2-25? amu atom " " amu/atom. The atomic weight of magnesium is amu/atom. Magnesium, Mg, is the element with an atomic weight slightly over amu (a) amu a measurement of mass that is equal to eactly 1/12 of the mass of an atom of carbon-12. (b) The mass of an atom of ruthenium is almost twice that of an atom of vanadium (101.07/50.94) Here we use the atomic weights to the number of places given in the periodic table in the inside front cover of the tet. 2-2

3 (a) 1 Ca = amu = amu 1 S = amu = amu 4 O = amu = amu FW = amu (b) 3 C = amu = amu 8 H = amu = amu FW = amu (c) 6 C = amu = amu 8 H = amu = amu 1 S = amu = amu 2 O = amu = amu 2 N = amu = amu FW = amu (d) 3 U = amu = amu 14 O = amu = amu 2 P = amu = amu 2-31 The ratio of masses present is ratio represents 6.68 g Ca 6.33 g F FW = amu = or g Ca 1.0 g F. Based on the formula CaF 2, this 1 atom Ca AW Ca. So the atomic mass ratio of Ca/F is 2 atoms F AW F From a table of atomic weights, AW Ca AW F = amu amu = 2.11 or /2 = 2.11 The first calculation could not be done without knowledge of the formula or some other knowledge of the relative numbers of atoms present. 2-33? g H 2 O 2 = 1.24 mol H 2 O g H 2 O 2 1 mol H 2 O 2 = 42.2 g H 2 O (a)? Formula Units K 2 = g K 2 1 mol K g K 2 (b)? K + ions = Formula Units K For. Units K 2 1 mol K 2 = Form. Units K K + ions 1 For. unit K 2 = K + ions

4 (c)? 2 ions = Formula Units K ion 1 For. Unit K 2 = (d) Each formula unit contains 2 K, 1 Cr, and 4 O atoms, or 7 atoms total.? atoms = Formula units K2CrO ions 7 atoms 1 For. Unit K 2 = atoms g Ne g Ne per mole = mole Ne 2-39 (a) No. The molecular formulas are different, so the mass of one mole of molecules (the molar mass) is different. (b) Yes. One mole of any kind of molecules contains Avogadro s number of molecules. (c) No. (d) No. This is for the same reason given in (a). The formulas are different, so there are different numbers of atoms per molecule and, hence, different total numbers of atoms in equal numbers of molecules Here we show values in the table on the right front inside cover. The bolded amounts represents the amounts the students fill in. Formula Mass of one mole of molecules (a) Br Br g (b) O O g (c) P P g (d) Ne Ne g (e) S S g (f) O O g 2-43? g/atom Cu = g Cu 1 mol Cu 1 mol Cu atoms Cu = g/ 1 atom Cu 2-45? molecules C 3 H 8 = molecules CH 4 1 mol CH 4 1 mol C 3 H FW Fe 3 (PO 4 ) 2 = amu % Fe = amu Fe amu mol CH molecules CH g CH 4 1 mol CH 4 1 mol C 3 H g C 3 H molecules C 3 H 8 mol C 3 H 8 = molecules C 3 H 8 100% = 46.8% Fe

5 2-49 Mass of Moles of C = mol = 3.00 H = mol = 7.94 O = mol = 1.00 Whole-Number Ratio is C 3 H 8 O, the simplest formula. Formula weight of simplest formula = 60 amu. Since the formula weight of the simplest formula (FW = amu) is equal to the approimate molecular weight given, the molecular formula is the simplest formula, C 3 H 8 O 2-51 (a) % O = 100 % total [9.79% H % C] = 11.09% O So, MW = amu = amu (b) % O = 100 % total [9.79% H % C] = 11.09% O Rel. Mass Multiply by 2 C = = H = = O = = The simplest formula is C 19 H 28 O 2. Given that each molecule contains two O atoms, the molecular formula is C 19 H 28 O 2. As a check on the MW calculated above, the MW of this formula is

6 2-53 (a) (b) 2-55 (a) Rel. Mass Cu = = 1.00 C = = 4.00 H = = 4.01 O = = 6.00 Rel. Mass The simplest formula is CuC 4 H 4 O 6 N = = 1.00 O = * = 1.00 B = = 1.00 F = = 4.00 The simplest formula is NOBF 4 *More significant digits can be kept throughout the problem and rounded for the final answer. Mass of N = = Cl = = 1.00 H = = The simplest formula is NClH 2 or NH 2 Cl 2-6

7 (b) Rel. Mass N = = 1.00 Cl = = 1.00 H = 7.43* = The simplest formula is NClH 4 or NH 4 Cl *More significant digits can be kept throughout the problem and rounded for the final answer Rel. Mass C H Cl N O = = 7.51 = = = = = = 1.00 = 1.00 = 1.00 The simplest formula is C 13 H 18 ClNO 2-7

8 2-59 Rel. Mass C = = H = = O = = 4.01 N = = 1.00 The simplest formula is C 17 H 21 O 4 N 2-61 (a) FW C 14 H 18 N 2 O 5 = amu % C = % H = % N = % O = amu C amu amu H amu amu N amu amu O amu 100% = 57.13% C 100% = 6.18% H 100% = 9.520% N 100% = 27.18% O (b) FW SiC = amu % Si = % C = amu Si amu amu C amu (c) FW C 9 H 8 O 4 = amu 100% = 70.05% Si 100% = 29.95% C % C = % H = % O = amu C amu amu H amu amu O amu 100% = 59.99% C 100% = 4.48% H 100% = 35.52% O 2-8

9 2-63 (a) Hydrogen peroide s actual formula is H 2 O 2 ; however, its simplest formula or lowest whole number ratio is HO. (b) Water s actual formula is H 2 O, while its simplest formula is also H 2 O. (c) Ethylene glycol s actual formula is C 2 H 6 O 2 ; however, its simplest formula is CH 3 O. 2-65?g C = 2.92 g CO 2?g H = 1.22 g H 2 O g C g CO 2 = g C 2(1.008 g H) g H 2 O = g H?g O = 1.20 g g g = 0.27 g O Mass of C H O = = = = = = 1.00 The simplest formula is C 4 H 8 O 2-67?mol C = g CO g C g CO 2 = g C?g H = g H 2 O 2( g H) g H 2 O = g H?g N = g ( ) = g N Mass of C H = = = = N = = 1.00 The simplest formula is CH 4 N, which has a molar mass of g/mol The actual substance has a molar mass of g/mol The molecular formula is C 2 H 8 N 2, = 2 2-9

10 2-69? g Mg = g MgO 24.3 g Mg g MgO = g Mg? g H = g H 2 O g H g H 2 O = g H? g Si = g SiO g Si 60.1 g SiO 2 = g Si? g O = g total [ g Mg g H g Si] = g O Mass of Mg = = 1.00 H = = 1.00 Si = = 1.00 O = = The simplest formula is MgHSiO Calculate the amount of O for a given amount of H in each compound: In H 2 O: In H 2 O 2 : amu O amu H amu O amu H = 7.92 amu O/amu H = amu O/amu H The mass of O in these two compounds is in the ratio 7.92 : or 1 : 2. The masses of O that combine with a fied mass of H in the two compounds are in the ratio of small whole numbers, 1 : 2. Alternatively, the masses of H that combine with a fied mass of O could be compared If the M 2 O substance is 73.4% M by mass, then it is 26.6% Oygen by mass. This means that if you had one mole of M 2 O: 26.6 = g O g M 2 O 100 or that would be the grams of M 2 O in a mole = as the mass of the 2 M atoms; each M is g/mol So for MO:? % M in MO = g M g MO 100 = 58.0 % M in MO 2-10

11 2-75 Note: The mass (or weight) ratio in any units is the same as that deduced in amus or grams, e.g., amu Cu amu CuFeS 2 or? lb Cu = 5.82 lb CuFeS (a) FW CuSO 4 = amu? g Cu = 573 g CuSO 4 (b) FW CuSO 4 5H 2 O = lb Cu lb CuFeS lb Cu lb CuFeS 2 = 2.02 lb Cu g Cu g CuSO 4 = 228 g Cu? g Cu = 573 g CuSO 4 5H 2 O g Cu g CuSO 4 5H 2 O = 146 g Cu 2-79? g Cu 3 (CO 3 ) 2 (OH) 2 = 378 g Cu g Cu 3 (CO 3 ) 2 (OH) g Cu 2-81 Formula weights: CaWO 4 = ; FeWO 4 = ? g CaWO 4 = 657 g FeWO 4 = g Cu 3 (CO 3 ) 2 (OH) g W g FeWO g CaWO g W = 623 g CaWO ? g Pb = 205 g ore 10.0 g PbS g ore g Pb g PbS = 17.8 g Pb/205 g ore 2-85? g Sr(NO 3 ) 2 = g sample 88.2 g Sr(NO 3) g sample = 278 g Sr(NO 3) 2 present The formula weight of Sr(NO 3 ) 2 is g/mol. (a)? g Sr = 278 g Sr(NO 3 ) 2 (b)? g N = 278 g Sr(NO 3 ) g Sr g Sr(NO 3 ) 2 = 115 g Sr g N g Sr(NO 3 ) 2 = 36.8 g N 2-87 (a)? g CH 3 COOH = g vinegar 5.0 g CH 3COOH 100 g vinegar = 11 g CH 3 COOH (b)? lb CH 3 COOH = lb vinegar 5.0 lb CH 3COOH 100 lb vinegar = 11 lb vinegar (c)? g NaCl = 51.9 g solution 5.0 g NaCl 100 g solution = 2.6 g NaCl 2-11

12 2-89 Assume you spend one dollar to purchase each substance. To get the lb of nitrogen per dollar:? lb NH 4 NO 3 $? lb CH 4 N 2 O $ = 1 lb NH 4 NO 3 $ 2.95 = 1 lb CH 4 N 2 O $ 3.65 CH 4 N 2 O has more N for the dollar lb N lb NH 4 NO 3 = lb N per dollar for NH 4 NO lb N lb CH 4 N 2 O = lb N per dollar for CH 4 N 2 O 2-91 The chemical formula for calcium carbonate is CaCO 3, and its molar mass is g/mol. The mass of CaCO 3 needed to supply 1200 mg of Ca per day =? g CaCO 3 /day = 1200 mg Ca 1 day 2-93 Let = atomic weight of metal M. (a) % M = 52.9% = 1 g Ca 1000 mg Ca g CaCO g Ca mass M mass M + mass O 100% = (3 AW O) 100% = = 27.0 amu ( ) 100 ; = 2 ; = (b) The metal is probably aluminum (atomic weight 26.98) MW = g/mol or amu/molecule = 3.0 g CaCO 3 /day? Fe atoms molecule = amu hemoglobin 0.35 amu Fe 1 molecule 100 amu hemoglobin 1 Fe atom amu Fe = 4.1 There are 4 iron atoms per hemoglobin molecule FW Ca 10 (PO 4 ) 6 (OH) 2 = amu (a) % Ca = (b) % P = amu Ca amu amu P amu 100% = 39.89% Ca 100% = 18.50% P 2-99 For the cations given, the group number is the same as the charge. Rubidium would likely form a 1+ ion since it is in group 1. The formula for the cation formed from the barium atom would be: Ba 2+. The formula for the anion formed from the nitrogen atom would be: N The new and old values for Avogadro's number are the same up to 7 significant digits; both are equal to , but differ in the net digit. The uncertainty only has 2 significant digits (1.5

13 10 17 ). If the uncertainty were subtracted from , the result would be , so with the uncertainty, the two numbers are the same to 7 significant digits ( and 602,214,141,070,409,084,099,072) All have the same empirical formula, CH 2 O, which has a formula weight (FW) of 30.0 amu. Molecular Molecular Ratio With Formula Weight (amu) Empirical FW Acetic Acid C 2 H 4 O amu 2 Erythrulose C 4 H 8 O amu 4 Formaldehyde CH 2 O 30.0 amu 1 Latic Acid C 3 H 6 O amu 3 Ribose C 5 H 10 O amu There is insufficient information since the oygen used in combustion comes from the air in addition to the oygen in the sample % Ag in Ag 2 O = % Ag in Ag 2 S = g Ag g Ag 2 O g Ag g Ag 2 S 100% = 93.10% Ag 100% = 87.06% Ag Recommend that, if the ores are the same price and if they contain the same mass percent of the silver compounds, the silver oide be used. However, in an actual situation, the price and concentration of the desired compound would probably be the determining factors. Pure Ag 2 O and Ag 2 S both contain a very high percentage of silver picomole mole 1 picomole pennies 1 mole 1 16 in penny 1 ft 12 in 1 mile 5280 ft = miles which is greater than 222,000 miles Yes, it will reach the moon ? MW or g/mol of B 12 = 100 g B g Co g Co 1 mol Co 1mol Co 1 mol B12 = g/mol B

14 2-113 % O = 100 % total 92.83% Pb = 7.17% O Rel. Mass Pb = = 1.00 O = = ml H 2 O = cm 3 H 2 O, since 1 ml = 1cm 3 The simplest formula is PbO? molecules of H 2 O = cm 3 H 2 O 1 mol H 2 O g H 2 O molecules H2 O 1 mol H 2 O = molecules H 2 O 2-14

Formulas and Models 1

Formulas and Models 1 A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance An empirical formula shows the simplest whole-number ratio of the atoms in