Respuesta :
Answer:
The empirical formula of compound X [tex]C_3H_{6}O_1[/tex].
The molecular formula of compound X [tex]C_6H_{12}O_2[/tex].
Explanation:
Compound X contains 62% carbon and 10.4% hydrogen.Suppose in 100 g of compound 62 grams of carbon ,10.4 grams of hydrogen is present and 27.6 grams.
Moles of carbon = [tex]\frac{62 g}{12 g/mol}=5.17 mol[/tex]
Moles of hydrogen = [tex]\frac{10.4 g}{1 g/mol}=10.4 mol[/tex]
Moles of oxygen = [tex]\frac{27.6 g}{16 g/mol}=1.725 mol[/tex]
For empirical formula divide the smallest number of moles of from all the moles of element:
For carbon = [tex]\frac{5.17 mol}{1.725 mol}=3[/tex]
For hydrogen = [tex]\frac{10.4mol}{1.725 mol}=6[/tex]
For oxygen = [tex]\frac{1.725 mol}{1.725 mol}=1[/tex]
Empirical formula for compound X :[tex]C_3H_6O[/tex]
The empirical formula of compound X [tex]C_3H_{6}O_1[/tex].
Empirical mass for compound X :
=[tex]3\times 12 g/mol+6\times 1g/mol+1\times 16 g/mol =58 g/mol[/tex]
Molecular mass of the compound X = 117 g/mol
[tex]\text{Moleculare mass}=n\times \text{Empirical mass}[/tex]
[tex]117 g/mol=n\times 58 g/mol[/tex]
[tex]n=\frac{117 g/mol}{58 g/mol}=2.0[/tex]
Molecular formula for compound X :[tex]C_{n\times 3}H_{n\times 6}O_{n\times 1}=C_6H_{12}O_2[/tex]
The molecular formula of compound X [tex]C_6H_{12}O_2[/tex].
