Question

When 3.62 g of a compound containing carbon, hydrogen, and oxygen were burned completely in air, 5.19 g of CO2 and 2.83 g of H2O were produced. What is the empirical formula of the compound?

Answer 1

Answer on Question #48772, Chemistry, Other

Task:

When 3.62 g of a compound containing carbon, hydrogen, and oxygen were burned completely in air, 5.19 g of CO₂ and 2.83 g of H₂O were produced. What is the empirical formula of the compound?

Answer:

$$\mathrm{C}_x\mathrm{H}_y\mathrm{O}_z + (x + y/4 - z/2)\mathrm{O}_2 = x\mathrm{CO}_2 + y/2\mathrm{H}_2\mathrm{O}$$ $$\nu = \frac{m}{M}$$ $$M(\mathrm{CO}_2) = 44\,g/mol$$ $$\nu(\mathrm{CO}_2) = \frac{5.19}{44} = 0.118\,mol$$ $$\nu(\mathrm{C}) = 0.118\,mol$$ $$m(\mathrm{C}) = \nu M = 0.118 \cdot 12 = 1.41\,g$$ $$M(\mathrm{H}_2\mathrm{O}) = 18\,g/mol$$ $$\nu(\mathrm{H}_2\mathrm{O}) = \frac{2.83}{18} = 0.16\,moles$$ $$\nu(\mathrm{H}) = 0.16 \cdot 2 = 0.32\,moles$$ $$m(\mathrm{H}) = \nu M = 0.32 \cdot 1 = 0.32\,g$$ $$m(\mathrm{O}) = 3.62 - 1.41 - 0.32 = 3.2\,g$$ $$\nu(\mathrm{H}) = \frac{3.2}{16} = 0.118\,moles$$

C:H:O=0.118:0.32:0.118

C:H:O=1:2.66:1

After multiplying by 3 we will receive: C:H:O=3:8:3

The final formula is: $\mathrm{C}_3\mathrm{H}_8\mathrm{O}_3$

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