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Mass Relationships in Chemical Reactions: Calculating Molar Masses and Balanced Equations, Study notes of Chemistry

Examples of calculating molar masses of various chemical compounds and balancing chemical equations. It includes step-by-step calculations for various chemical formulas and molecular masses.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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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
3.13
23
6.022 10 S atoms
5.10 mol S 1molS
×
×=
24
3.07 10 S atoms×
3.15 1mol Ca
77.4 g of Ca 40.08 g Ca
×=1.93 mol Ca
3.23 Using the appropriate atomic masses,
(a) CH4 12.01 amu + 4(1.008 amu) = 16.04 amu
(b) NO2 14.01 amu + 2(16.00 amu) = 46.01 amu
(c) SO3 32.07 amu + 3(16.00 amu) = 80.07 amu
(d) C6H6 6(12.01 amu) + 6(1.008 amu) = 78.11 amu
(e) NaI 22.99 amu + 126.9 amu = 149.9 amu
(f) K2SO4 2(39.10 amu) + 32.07 amu + 4(16.00 amu) = 174.27 amu
(g) Ca3(PO4)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 C19H38O 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 × 1012 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
910 910
910
910 910
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 × 1021 molecules C9H10O
pf3

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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 )