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CHAPTER 3
MASS RELATIONSHIPS IN
CHEMICAL REACTIONS
3.5 (34.968 amu)(0.7553) + (36.956 amu)(0.2447) = 35.45 amu
23 6.022 10 S atoms 5.10 mol S 1 mol S
×
× =
24 3.07 × 10 S atoms
1 mol Ca 77.4 g of Ca 40.08 g Ca
× = 1.93 mol Ca
3.23 Using the appropriate atomic masses,
(a) CH 4 12.01 amu + 4(1.008 amu) = 16.04 amu
(b) NO 2 14.01 amu + 2(16.00 amu) = 46.01 amu
(c) SO 3 32.07 amu + 3(16.00 amu) = 80.07 amu
(d) C 6 H 6 6(12.01 amu) + 6(1.008 amu) = 78.11 amu
(e) NaI 22.99 amu + 126.9 amu = 149.9 amu
(f) K 2 SO 4 2(39.10 amu) + 32.07 amu + 4(16.00 amu) = 174.27 amu
(g) Ca 3 (PO 4 ) 2 3(40.08 amu) + 2(30.97 amu) + 8(16.00 amu) = 310.18 amu
3.25 To find the molar mass (g/mol), we simply divide the mass (in g) by the number of moles.
152 g
0.372 mol
= 409 g/mol
3.29 The molar mass of C 19 H 38 O is 282.5 g.
23 12 1 mol^ 6.022^10 molecules 1.0 10 g 282.5 g 1 mol
− ×
× × × =
9 2.1 × 10 molecules
Notice that even though 1.0 × 10
− 12 g is an extremely small mass, it still is comprised of over a billion pheromone molecules!
3.41 The molar mass of cinnamic alcohol is 134.17 g/mol.
(a)
(9)(12.01 g/mol) 100% 134.17 g/mol
%C = × = 80.56%
(10)(1.008 g/mol) 100% 134.17 g/mol
%H = × = 7.51%
16.00 g/mol 100% 134.17 g/mol
%O = × = 11.93%
(b)
23 9 10 9 10 9 10 9 10 9 10
1 mol C H O 6.022 10 molecules C H O 0.469 g C H O 134.17 g C H O 1 mol C H O
×
× ×
= 2.11 × 10
21 molecules C 9 H 10 O
2 CHAPTER 3: MASS RELATIONSHIPS IN CHEMICAL REACTIONS
3.43 Assume you have exactly 100 g of substance.
C
1 mol C 44.4 g C 3.70 mol C 12.01 g C
n = × =
H
1 mol H 6.21 g H 6.16 mol H 1.008 g H
n = × =
S
1 mol S 39.5 g S 1.23 mol S 32.07 g S
n = × =
O
1 mol O 9.86 g O 0.616 mol O 16.00 g O
n = × =
Thus, we arrive at the formula C3.70 H (^) 6.16 S (^) 1.23 O (^) 0.616. Dividing by the smallest number of moles (0.616 mole)
gives the empirical formula, C 6 H 10 S 2 O.
To determine the molecular formula, divide the molar mass by the empirical mass.
molar mass 162 g 1 empirical molar mass 162.3 g
Hence, the molecular formula and the empirical formula are the same, C 6 H 10 S 2 O.
3.52 The empirical molar mass of CH is approximately 13.02 g. Let's compare this to the molar mass to determine
the molecular formula.
Recall that the molar mass divided by the empirical mass will be an integer greater than or equal to one.
molar mass 1 (integer values) empirical molar mass
In this case,
molar mass 78 g 6 empirical molar mass 13.02 g
Thus, there are six CH units in each molecule of the compound, so the molecular formula is (CH) 6 , or C 6 H 6.
3.59 The balanced equations are as follows:
(a) 2C + O 2 → 2CO (h) N 2 + 3H 2 → 2NH (^3)
(b) 2CO + O 2 → 2CO 2 (i) Zn + 2AgCl → ZnCl 2 + 2Ag
(c) H 2 + Br 2 → 2HBr (j) S 8 + 8O 2 → 8SO (^2)
(d) 2K + 2H 2 O → 2KOH + H 2 (k) 2NaOH + H 2 SO 4 → Na 2 SO 4 + 2H 2 O
(e) 2Mg + O 2 → 2MgO (l) Cl 2 + 2NaI → 2NaCl + I (^2)
(f) 2O 3 → 3O 2 (m) 3KOH + H 3 PO 4 → K 3 PO 4 + 3H 2 O
(g) 2H 2 O 2 → 2H 2 O + O 2 (n) CH 4 + 4Br 2 → CBr 4 + 4HBr
3.67 Starting with the amount of ammonia produced (6.0 moles), we can use the mole ratio from the balanced
equation to calculate the moles of H 2 and N 2 that reacted to produce 6.0 moles of NH 3.
3H 2 ( g ) + N 2 ( g ) → 2NH 3 ( g )
