Question

A sample containing only carbon, hydrogen, and silicon is subjected to elemental analysis. After complete combustion, a 0.7020 g sample of the compound yields 1.4 g of CO2, 0.86 g of H2O, and 0.478 g of SiO2. What is the empirical formula of the compound?

Answer 1

Answer: The empirical formula of compound is
`C_4H_(12)Si`.

Step-by-step explanation:

Mass of Sample= 0.702 g

Mass of
`CO_2` = 1.4 g

Mass of
`H_2O` = 0.86 g

Mass of
`SiO_2` = 0.478 g

First we have to calculate moles of
`CO_2`,
`H_2O` and
`SiO_2` formed.

1. Moles of
`CO_2=(1.4g)/(44g/mol)=0.032mol`

Now , Moles of carbon == Moles of
`CO_2` = 0.032

2. Moles of
`H_2O=(0.86g)/(18g/mol)`=0.048mol​​​

Now , Moles of hydrogen =
`2*` Moles of
`H_2O` =
`2* 0.048=0.096mol`

3. Moles of
`SiO_2=(0.478g)/(60g/mol)=0.008` mol

Now , Moles of silicon = Moles of
`SiO_2` = 0.008 moles

Therefore, the ratio of number of moles of C : H : Si is = 0.032 : 0.096 : 0.008

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C=
`(0.032)/(0.008)=4`

For H =
`(0.096)/(0.008)=12`

For Si=
`(0.008)/(0.008)=1`

Thus, C: H: Si = 4 : 12 : 1

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is
`C_4H_(12)Si`.