Question
Asked Oct 3, 2019
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A sample of a substance (containing only C, H, and N) is burned in oxygen.
3.382 g of CO22.307×10-1 g of H2O and 7.685×10-1 g of NO are the sole products of combustion.
What is the empirical formula of the compound?

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Expert Answer

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Step 1

To calculate the empirical formula, it is required to calculate the amount of carbon, hydrogen and nitrogen present in the sample. This can be done as,

Mass
No. of moles of Carbon dioxide
Molar Mass
3.382
0.0768 moles
No. of moles of Carbon dioxide
44
In 1 mole of CO2, 12 grams of carbon is present,
Therefore in 0.0768 moles of CO2 = 12 x 0.0768 = 0.9216 gm of carbon will be
present
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Mass No. of moles of Carbon dioxide Molar Mass 3.382 0.0768 moles No. of moles of Carbon dioxide 44 In 1 mole of CO2, 12 grams of carbon is present, Therefore in 0.0768 moles of CO2 = 12 x 0.0768 = 0.9216 gm of carbon will be present

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Step 2

Calculating the mass of Hydrogen,

Mass
No. of moles of water
Molar Mass
0.2307
0.0128 moles
No. of moles of water
18
In 1 mole of H2O, 2 grams of H2 is present,
Therefore in 0.0128 moles of H20 = 0.0128 x 2 0.0256 gm of hydrogen will be
present
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Mass No. of moles of water Molar Mass 0.2307 0.0128 moles No. of moles of water 18 In 1 mole of H2O, 2 grams of H2 is present, Therefore in 0.0128 moles of H20 = 0.0128 x 2 0.0256 gm of hydrogen will be present

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Step 3

Calculating the mass...

No. of moles of NOMolar Mass
Mass
No. of moles of NO = 0.7685
30
No. of moles of NO 0.0256 moles
In 1 mole of NO, 14 grams of nitrogen is present,
Therefore in 0.0256 moles of NO 0.0256 x 14 0.358 gm of nitrogen will be
present.
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No. of moles of NOMolar Mass Mass No. of moles of NO = 0.7685 30 No. of moles of NO 0.0256 moles In 1 mole of NO, 14 grams of nitrogen is present, Therefore in 0.0256 moles of NO 0.0256 x 14 0.358 gm of nitrogen will be present.

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