Question

Combustion of 2.78 g of ethyl butyrate leads to the formation of 6.32 g of CO2 and 2.58 g of H2O. The Molar mass is between 100 and 150. What is the empirical and molecular formula?

Answer 1

Answer on Question#57210 – Chemistry – General chemistry

Question: Combustion of 2.78 g of ethyl butyrate leads to the formation of 6.32 g of CO₂ and 2.58 g of H₂O. The Molar mass is between 100 and 150. What is the empirical and molecular formula?

Solution:

1. Ethyl butyrate is compound that contains C, H and perhaps O. Check the presence of O:

$$m(O_2) = m(CO_2) + m(H_2O) - m(\text{ethyl butyrate}) = 6.32\,g + 2.58\,g - 2.78\,g = 6.12\,g.$$

In $CO_2$:

$$w(O) = \frac{M(O)}{M(CO_2)} = \frac{2 \cdot 16\,g/mol}{44\,g/mol} = 0.7272$$ $$m_1(O) = w(O) \cdot m(CO_2) = 0.7272 \cdot 6.32\,g = 4.5959\,g$$

In $H_2O$:

$$w(O) = \frac{M(O)}{M(H_2O)} = \frac{16\,g/mol}{18\,g/mol} = 0.8889$$ $$m_2(O) = w(O) \cdot m(H_2O) = 0.8889 \cdot 2.58 = 2.2934\,g$$ $$m_1(O) + m_2(O) = 4.5959\,g + 2.2934\,g = 6.89\,g$$ $$m_1(O) + m_2(O) > m(O_2) - \text{so our compound contains O}$$

2. Find the empirical formula:

$$n(CO_2) = \frac{6.32\,g}{44\,g/mol} = 0.1436\,mol$$ $$n(H_2O) = \frac{2.58\,g}{18\,g/mol} = 0.1433\,mol$$ $$n(O_2) = \frac{6.12\,g}{32\,g/mol} = 0.19125\,mol$$

In ethyl butyrate:

$$n(C) = n(CO_2) = 0.1436\,mol$$ $$n(H) = n(H_2O) = 0.2866\,mol$$ $$n(O) = n(H_2O) + 2n(CO_2) - 2n(O_2) = 0.1433\,mol + 0.2872\,mol - 0.3825\,mol = 0.048\,mol$$ $$N(C) : N(H) : N(O) = 0.1436 : 0.2866 : 0.048 = 3 : 6 : 1$$

So $C_3H_6O$ is empirical formula of ethyl butyrate.

3. Find the molecular formula. Molecular formula of ethyl butyrate is $(C_3H_6O)_n$

$$M(C_3H_6O) = 58\,g/mol$$ $$n_{\min} = \frac{100}{58} = 1.72$$ $$n_{\max} = \frac{150}{58} = 2.59$$

The only integer is 2. So molecular formula is $C_6H_{12}O_2$.

**Answer:** empirical formula is $C_3H_6O$

molecular formula is $C_6H_{12}O_2$.

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