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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 :

Answer: The empirical formula of compound is [tex]C_4H_{12}Si[/tex].

Explanation:

Mass of Sample= 0.702 g

Mass of [tex]CO_2[/tex] = 1.4 g

Mass of [tex]H_2O[/tex] = 0.86 g

Mass of [tex]SiO_2[/tex] = 0.478 g  

First we have to calculate moles of[tex]CO_2[/tex], [tex]H_2O[/tex] and [tex]SiO_2[/tex] formed.

1. Moles of [tex]CO_2=\frac{1.4g}{44g/mol}=0.032mol[/tex]

Now , Moles of carbon == Moles of [tex]CO_2[/tex] = 0.032

2.  Moles of [tex]H_2O=\frac{0.86g}{18g/mol}[/tex]=0.048mol​​​

Now , Moles of hydrogen = [tex]2\times[/tex] Moles of [tex]H_2O[/tex] =[tex]2\times 0.048=0.096mol[/tex]

3.  Moles of [tex]SiO_2=\frac{0.478g}{60g/mol}=0.008[/tex] mol

Now , Moles of silicon = Moles of [tex]SiO_2[/tex] = 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= [tex]\frac{0.032}{0.008}=4[/tex]

For H =[tex]\frac{0.096}{0.008}=12[/tex]

For Si=[tex]\frac{0.008}{0.008}=1[/tex]

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

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is [tex]C_4H_{12}Si[/tex].

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