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A 1.50 gram sample of contain 3.32g of CO2 1.58g of N2O5 and 1.865g of H2O .Its molar mass is 102.2g/mol. Determine the emperical and molecular formulas.

Sagot :

The correct answer for emperical and molecular formula = C5H14N2 .

3.23 g x (12.011 / 44.0098) = 0.8815 g carbon

1.865 g x (2.016 / 18.0152) = 0.2087 g Hydrogen

1.58 g x (28.014 / 108.009) = 0.4098 g Nitrogen

1.50 g minus (0.8815 + 0.2087 + 0.4098) = 0.

Convert each element's mass to moles.

0.8815 g/12.011 g/mol = 0.0734 mol carbon

0.2087 g 1.008 g/mol = 0.207 mol Hydrogen

0.4098 g 14.007 g/mol = 0.02926 mol Nitrogen

Step three is to calculate the ratio of molar amounts expressed in the smallest, whole numbers.

0.0734 mol 0.02926 mol = 2.51 carbon

7.07 mol hydrogen = 0.207 mol 0.02926 mol

Nitrogen: 0.02926 mol = 1 Nitrogen: 0.02926 mol = 1

Doubling each value yields C = 5, H = 14.14, and N = 2, resulting in the empirical formula C5H14N2

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