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A 1.638 Gram pure sample of a compound containing only carbon hydrogen and oxygen was burned in excess oxygen gas 3.117 g of carbon dioxide and 1.911 g of water were produced find the empirical formula of the compound

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dumbchild123

Answer:

C2H6O

Explanation:

First, find the masses of C, H, and O. I used ratios.

3.117g CO2  x  1 mol CO2/44.01g CO2  x  1 mol C/1 mol CO2  x  12.01g C/1 mol C = 0.850606 g C

1.911g H2O  x  1 mol H2O/18.016g H2O  x  2 mol H2O/1 mol H2O  x  1.008g H/1 mol H = 0.213842 g H

To find the mass of oxygen, just add the masses of C and H together and subtract that from the total mass of the sample.

1.638 - (0.850606 + 0.213842) = 0.57355 g O

Second, find the moles of C, H, and O.

0.850606g C  x  1 mol C/12.01g C = 0.070825 mol C

0.213842g H  x  1 mol H/1.008g H = 0.212145 mol H

0.57355 g O  x  1 mol O/16g O = 0.035847 mol O

Finally, divide the number of moles of each element by the smallest value of moles calculated. In this case, the smallest value is 0.035847.

0.070825/0.035847 = 2

0.212145/0.035847 = 6

0.035847/0.035847 = 1

The empirical formula represents the chemical with the ratio of the simplest whole numbers. The empirical formula of the 1.638-gram compound is C₂H₆O.

What is the empirical formula?

The representation of the chemical compound with the chemical symbols and the coefficients as the whole number are called empirical formulas.

The molecular formula for the sample is, CxHyOz

Given,

Mass of carbon dioxide = 3.117 gm

Mass of water = 1.911 gm

Mass of pure sample = 1.638 gram

The moles of carbon (x) are calculated as,

Moles = mass ÷ molar mass

= 3.117 ÷ 44.01

= 0.0708248

Moles of hydrogen (y) are calculated as,

Moles = 1.9911 ÷ 18.015

= 0.106078 moles

In water there are 2 hydrogen atoms so, 0.212156 moles

Mass of CxHyOz is used to calculate moles of oxygen as,

0.0708248 × 12.0107 + 0.212156 × 1.00784 + z × 15.999 = 1.638

1.0644747 + 15.999 z = 1.638

15.999 z = 0.5735253

z = 0.0358475

So, x = 0.0708248, y = 0.212156  and z = 0.0358475

These values are divided by the least number of moles to give the ratio as,

0.070825 ÷ 0.035847 = 2

0.212145 ÷ 0.035847 = 6

0.035847 ÷ 0.035847 = 1

Therefore, the empirical formula for the compound is C₂H₆O.

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