Question

(a) How many moles of ammonium ions are in 0.753 g of ammonium carbonate? mol

(b) What is the mass, in grams, of 0.0567 mol of iron(III) phosphate? g

(c) What is the mass, in grams, of 4.72 1023 molecules of aspirin, C9H8O4? g

(d) What is the molar mass of a particular compound if 0.050 mol weighs 5.44 g? g/mol

Answer 1

a) (NH4)2CO3 = 96.09 g/mol

moles = 0.753 / 96.09 = 0.0078

moles of ammonium = 2 x 0.0078  

moles of ammonium ion = 0.0156

b)

FePO4 = 150.82 g/mol

mass = moles x molar mass

mass = 0.0567 x 150.82

mass = 8.55 g

c)

C9H8O4 = 180.16 g/mol

6.023 x 10²³ molecules = 180.16 g

4.72 x 10²³ molecules = 4.72 x 10²³ x 180.16 / 6.023 x 10²³ = 141.18 g

mass = 141.18 g

d) moles = mass / molar mass

molar mass = mass / moles = 5..44 / 0.050

molar mass = 136 g/mol

Answer 2

Answers:

(a) Moles of ammonium ions in 0.753 g of ammonium carbonate: 

Answer: 0.0157 mol

Calculation:

1. Molar mass of ammonium carbonate, (NH₄)₂CO₃:

• N: 2 × 14.01 = 28.02 g/mol

• H: 8 × 1.01 = 8.08 g/mol

• C: 1 × 12.01 = 12.01 g/mol

• O: 3 × 16.00 = 48.00 g/mol

• 
  

• Total = 28.02 + 8.08 + 12.01 + 48.00 = 96.11 g/mol

2. Moles of (NH₄)₂CO₃ in 0.753 g:

   $$\text{Moles}=\frac{\text{Mass}}{\text{Molar Mass}}=\frac{0.753}{96.11}=0.00783\text{ mol}$$

4. Each (NH₄)₂CO₃ has 2 NH₄⁺ ions, so:

   $$\text{Moles of NH₄⁺}=0.00783\times 2=0.0157\text{ mol}$$

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(b) Mass of 0.0567 mol of iron(III) phosphate:
Answer: 8.53 g

Calculation:

1. Formula: FePO₄

2. Molar mass:

• Fe: 55.85 g/mol

• P: 30.97 g/mol

• O: 4 × 16.00 = 64.00 g/mol

• Total = 55.85 + 30.97 + 64.00 = 150.82 g/mol

3. Mass = Moles × Molar Mass:

   $$0.0567\times 150.82=8.53\text{ g}$$

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(c) Mass of 4.72 × 10²³ molecules of aspirin (C₉H₈O₄):
Answer: 141.26 g

Calculation:

1. Molar mass of aspirin:

• C: 9 × 12.01 = 108.09 g/mol

• H: 8 × 1.01 = 8.08 g/mol

• O: 4 × 16.00 = 64.00 g/mol

• Total = 108.09 + 8.08 + 64.00 = 180.17 g/mol

2. Moles from molecules (Avogadro’s number = 6.022 × 10²³):

   $$\text{Moles}=\frac{4.72\times 10^{23}}{6.022\times 10^{23}}=0.7838\text{ mol}$$

4. Mass = Moles × Molar Mass:

   $$0.7838\times 180.17=141.26\text{ g}$$

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(d) Molar mass of a compound (0.050 mol weighs 5.44 g):
Answer: 108.8 g/mol

Calculation:

$$\text{Molar Mass}=\frac{\text{Mass}}{\text{Moles}}=\frac{5.44}{0.050}=108.8\text{ g/mol}$$