Esters are common components of artificial flavours - VCE - SSCE Chemistry - Question 5 - 2002 - Paper 1

To find the mass of hydrogen, we start by calculating the number of moles of water produced.

1. Moles of $H_2O$:

   $\text{Moles of } H_2O = \frac{0.90 g}{18.02 g/mol} = 0.050 mol$

2. Each mole of water contains 2 moles of hydrogen, thus:

   $\text{Moles of H} = 2 \times 0.050 mol = 0.10 mol$

3. We calculate the mass of hydrogen:

   $\text{Mass of H} = 0.10 mol \times 1.01 g/mol = 0.10 g$

To find the mass of oxygen in the compound, we can use the total mass:

1. Total mass of the compound = 1.02 g

2. We have already determined:

   • Mass of C = 0.60 g
   • Mass of H = 0.10 g

3. The mass of oxygen is found by subtracting the mass of carbon and hydrogen from the total mass:

   $\text{Mass of O} = 1.02 g - (0.60 g + 0.10 g) = 0.32 g$

To find the empirical formula, we need to determine the ratio of moles of each element:

1. Moles of Carbon:

   $\text{Moles of C} = \frac{0.60 g}{12.01 g/mol} = 0.05 mol$

2. Moles of Hydrogen:

   $\text{Moles of H} = \frac{0.10 g}{1.01 g/mol} \approx 0.10 mol$

3. Moles of Oxygen:

   $\text{Moles of O} = \frac{0.32 g}{16.00 g/mol} = 0.02 mol$

4. The ratios are:

   • C: 0.05 mol
   • H: 0.10 mol
   • O: 0.02 mol

5. Divide each by the smallest number of moles:

   • C: $\frac{0.05}{0.02} = 2.5$
   • H: $\frac{0.10}{0.02} = 5$
   • O: $\frac{0.02}{0.02} = 1$

6. Since we can't have fractional subscripts, multiply each by 2:

   • C: 5
   • H: 10
   • O: 2

7. Therefore, the empirical formula is $C_5H_{10}O_2$.