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a 9.791 gram sample of an organic compound containing , and is analyzed by combustion analysis and 14.35 grams of and 5.876 grams of are produced. in a separate experiment, the molar mass is found to be 60.05 g/mol. determine the empirical formula and the molecular formula of the organic compound.

Answer :

The empirical formula of the compound is CH₂O, and the molecular formula of the compound is C₂H₄O₂.

All of the carbon in the compound is converted into carbon dioxide (14.35 g) and all the hydrogen in the compound is converted to water (5.876 g). This means that we can use these masses to find the masses of carbon, hydrogen, and oxygen in the original compound.

First, we calculate the mass of carbon in carbon dioxide and the mass of hydrogen in water, using their respective molar masses (M(CO₂) = 44 g/mol; M(H₂O) = 18 g/mol):

12 g of carbon : 44 g of carbon dioxide = X : 14.35 g of carbon dioxide

X = 12 g of carbon * 14.35 g of carbon dioxide / 44 g of carbon dioxide

X = 3.914 g of carbon

2 g of hydrogen : 18 g of water = X : 5.876 g of water

X = 2 g of hydrogen : 5.876 g of water / 18 g of water

X = 0.6529 g of hydrogen

Because the compound contains carbon, hydrogen and oxygen, we can now calculate the mass of oxygen in the compound:

m(O) = m(sample) - m(C) - m(H)

m(O) = 9.791 g - 3.914 g - 0.6529 g

m(O) = 5.224 g

Now we use these masses to calculate the number of moles (n) for each of the elements, using their respective molar masses (M(C) = 12 g/mol, M(H) = 1 g/mol, M(O) = 16 g/mol):

n = m/M

n(C) = m(C) / M(C)

n(C) = 3.914 g / 12 g/mol

n(C) = 0.3284 mol

n(H) = m(H) / M(H)

n(H) = 0.6529 g / 1 g/mol

n(H) = 0.6529 mol

n(O) = m(O) / M(O)

n(O) = 5.224 g / 16 g/mol

n(O) = 0.3265 mol

We divide each of these amounts by the smallest of them (0.3265 mol) to obtain the molar ratio of the elements:

carbon: 0.3284 mol / 0.3265 mol = 1.006

hydrogen: 0.6529 mol / 0.3265 mol = 2.000

oxygen: 0.3265 mol / 0.3265 mol = 1.000

These integers are indexes in the empirical formula of the compound - CH₂O

The molar mass of this empirical formula is:

12 g/mol + 2 * 1 g/mol + 16 g/mol = 30 g/mol

We divide the molar mass of the compound by the molar mass of the empirical formula:

60.05 g/mol / 30 g/mol = 2.002

So, we multiply the indexes by 2 to obtain the molecular formula - C₂H₄O₂

You can learn more about empirical formulas here:

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