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Olenka [21]
3 years ago
7

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

Chemistry
1 answer:
fiasKO [112]3 years ago
4 0

Answer: The empirical formula is CH_2.

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = \frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles

Moles of H = \frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.142}{0.142}=1

For H =\frac{0.287}{0.142}=2

The ratio of C: H = 1: 2

Hence the empirical formula is CH_2.

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