Question

What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

Answer 1

Answer:

$C_3H_5O_2$ the empirical formula of a compound.

Explanation:

Suppose in 100 g of compound contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen;

Then mass of carbon = $100\%\times \frac{50.0}{100}=50.0 g$

Moles of carbon =$\frac{50.0 g}{12 g/mol}=4.16 mol$

Then mass of hydrogen= $100\%\times \frac{6.7}{100}=6.7 g g$

Moles of hydrogen=$\frac{6.7 g}{1 g/mol}=6.70 mol$

Then mass of oxygen = $100\%\times \frac{43.3}{100}=43.3 g$

Moles of oxygen =$\frac{43.3 g}{16 g/mol}=2.70 mol$

To determine the empirical formula compound dived the smallest value of moles from each moles of element.

Carbon = $\frac{4.1666 mol}{2.7062 mol}=1.5$

Hydrogen= $\frac{6.70 mol}{2.7062 mol}=2.5$

Oxygen= $\frac{2.7062 mol}{2.7062 mol}=1$

The empirical formula ;$C_{1.5}H_{2.5}O_{1}=C_3H_5O_2$

$C_3H_5O_2$ the empirical formula of a compound.

Answer 2

In the layman's terms, air?
but from the list C3H2O5 looks right to me