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Chapter 3: Problem 24

Determine the formula weights of each of the following compounds: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) which is responsible for the rotten smell of spoiled food; (b) sodium perborate, \(\mathrm{NaBO}_{3}\), a substance used as bleach; (c) calcium carbonate, \(\mathrm{CaCO}_{3},\) a substance found in marble. (c) $\mathrm{CF}_{2} \mathrm{Cl}_{2},\( a refrigerant known as Freon; \)(\mathbf{d}) \mathrm{NaHCO}_{3},$ known as baking soda and used in bread and pastry baking; \((\mathbf{e})\) iron pyrite, \(\mathrm{FeS}_{2}\) which has a golden appearance and is known as "Fool's Gold."

Short Answer

Expert verified
The formula weights for the given compounds are as follows: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\): 88.12 (b) Sodium perborate, \(\mathrm{NaBO}_{3}\): 81.80 (c) Calcium carbonate, \(\mathrm{CaCO}_{3}\): 100.09 (d) Freon, \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\): 120.91 (e) Iron pyrite, \(\mathrm{FeS}_{2}\): 119.99

Step by step solution

01

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in butyric acid: - C: 12.01 - H: 1.01 - O: 16.00 Remember that there are 4 Carbon atoms, 8 Hydrogen atoms, and 2 Oxygen atoms in the formula.
02

Multiply and sum

Multiply the atomic weights by the number of atoms and then sum the results: Formula weight = (4 × 12.01) + (8 × 1.01) + (2 × 16.00) = 48.04 + 8.08 + 32.00 = 88.12 (b) Sodium perborate, \(\mathrm{NaBO}_{3}\)
03

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in sodium perborate: - Na: 22.99 - B: 10.81 - O: 16.00
04

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 22.99 + 10.81 + (3 × 16.00) = 22.99 + 10.81 + 48.00 = 81.80 (c) Calcium carbonate, \(\mathrm{CaCO}_{3}\)
05

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in calcium carbonate: - Ca: 40.08 - C: 12.01 - O: 16.00
06

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 40.08 + 12.01 + (3 × 16.00) = 40.08 + 12.01 + 48.00 = 100.09 (d) Freon, \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\)
07

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in Freon: - C: 12.01 - F: 19.00 - Cl: 35.45
08

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 12.01 + (2 × 19.00) + (2 × 35.45) = 12.01 + 38.00 + 70.90 = 120.91 (e) Iron pyrite, \(\mathrm{FeS}_{2}\)
09

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in iron pyrite: - Fe: 55.85 - S: 32.07
10

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 55.85 + (2 × 32.07) = 55.85 + 64.14 = 119.99

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