Question

1.)How much iron is present in 8 . 13 g of iron(III) oxide? Answer in units of g.

2.) A chemist wants to extract the gold from 62 . 21 g of AuCl 3 · 2 H 2 O (gold(III) chloride dihydrate) by electrolysis of an aqueous solu- tion. What mass of gold could be obtained from this sample? Answer in units of g.

3 (part 1 of 2) 10.0 points The molecular weight of erbium is 167 . 259 g/mol, sulfur is 32 . 065 g/mol, oxygen is 15 . 9994 g/mol, hydrogen is 1 . 00794 g/mol, carbon is 12 . 0107 g/mol, tin is 118 . 71 g/mol, and strontium is 87 . 62 g/mol. What is the percentage of C in glycerol? Answer in units of %.

4 (part 2 of 2) 10.0 points What is the percentage of H in glycerol? Answer in units of %.

5). 10.0 points What is the % carbon, by weight, in a 0 . 166 g sample of C 2 H 6 ? Answer in units of %.

Answer 1

Answer on the question #65594, Chemistry / Other

Question:

1.) How much iron is present in 8.13 g of iron(III) oxide? Answer in units of g.

2.) A chemist wants to extract the gold from 62.21 g of AuCl 3·2 H 2 O (gold(III) chloride dihydrate) by electrolysis of an aqueous solution. What mass of gold could be obtained from this sample? Answer in units of g.

3 (part 1 of 2) 10.0 points The molecular weight of erbium is 167.259 g/mol, sulfur is 32.065 g/mol, oxygen is 15.9994 g/mol, hydrogen is 1.00794 g/mol, carbon is 12.0107 g/mol, tin is 118.71 g/mol, and strontium is 87.62 g/mol. What is the percentage of C in glycerol? Answer in units of %.

4 (part 2 of 2) 10.0 points What is the percentage of H in glycerol? Answer in units of %.

5). 10.0 points What is the % carbon, by weight, in a 0.166 g sample of C 2 H 6? Answer in units of %.

Solution:

1) Iron (III) oxide formula is Fe₂O₃. Then, molar mass of Fe₂O₃ is 159.69 g/mol. The number of the moles is Fe₂O₃:

$$n(Fe_2O_3) = \frac{m}{M} = \frac{62.21(g)}{159.69(g \text{ mol}^{-1})} = 0.3896 \text{ mol}$$

The molar mass of iron is 55.845 g/mol. The number of the moles of iron and iron(III) oxide relate as:

$$n(Fe_2O_3) = \frac{n(Fe)}{2}$$

Then, the mass of iron is:

$$m(Fe) = n \cdot M = 0.3896(g) \cdot 2 \cdot 55.845(g \text{ mol}^{-1}) = 43.51g$$

2) As it can be seen from formula, the number of the moles of gold (molar mass of 196.966569 g/mol) and gold chloride dihydrate (molar mass of 339.3561 g/mol) are equal. So, the mass of gold in the compound is:

$$m(Au) = n \cdot M = \frac{62.21(g)}{339.3561(g \text{ mol}^{-1})} \cdot 196.966569(g \text{ mol}^{-1}) = 36.10(g)$$

3) The formula of glycerol is C₃H₈O₃, and the molar mass is 92.0938 g/mol. Also, we know that the molar mass of carbon is 12.0107 g/mol, and there are 3 atoms of carbon in one molecule of glycerol. Then, the mass percentage of carbon in glycerol is:

$$\omega(C) = \frac{m(C)}{m(C_3H_8O_3)} \cdot 100\% = \frac{3M(C)}{M(C_3H_8O_3)} \cdot 100\% = \frac{3 \cdot 12.0107(g \text{ mol}^{-1})}{92.0938(g \text{ mol}^{-1})} \cdot 100\%$$ $$\omega (C) = 39.1\%$$

4) The percentage of hydrogen in glycerol:

$$\omega (H) = \frac {m (H)}{m \left(C _ {3} H _ {8} O _ {3}\right)} \cdot 100 \% = \frac {8 M (H)}{M \left(C _ {3} H _ {8} O _ {3}\right)} \cdot 100 \% = \frac {8 \cdot 1.00794 (g \, mol^{-1})}{92.0938 (g \, m\partial l^{-1})} \cdot 100 \%$$ $$\omega (H) = 8.76\%$$

5) The mass percentage of carbon in $C_2H_6$ is:

$$\omega (C) = \frac {m (C)}{m \left(C _ {2} H _ {6}\right)} \cdot 100 \% = \frac {2 M (C)}{M \left(C _ {2} H _ {6}\right)} \cdot 100 \% = \frac {2 \cdot 12.0107 (g \, mol^{-1})}{30.0690 (g \, mol^{-1})} \cdot 100 \%$$ $$\omega (C) = 79.9\%$$

Answer: 1) $43.51 \text{ g}$ 2) $36.10 \text{ g}$ 3) $39.1\%$ 4) $8.76\%$ 5) $79.9\%$

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