A compound is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine. What is its empirical formula?

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A compound is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine. What is its empirical formula?

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Answer 1 · 2017-04-29T18:17:51

As with these problems, we assume a mass of $100 \cdot g$ $100*g$ of unknown compound..........and get an empirical formula of $C H_{3} B r$ $CH_3Br$

Explanation:

If there is a $100 \cdot g$ $100*g$ mass of compound, then we interrogate its ATOMIC composition......

$Moles of carbon = \frac{12.7 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}} = 1.06 \cdot m o l$ $"Moles of carbon"=(12.7*g)/(12.011*g*mol^-1)=1.06*mol$ .

$Moles of hydrogen = \frac{3.2 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}} = 3.175 \cdot m o l$ $"Moles of hydrogen"=(3.2*g)/(1.00794*g*mol^-1)=3.175*mol$ .

$Moles of bromine = \frac{84.1 \cdot g}{79.90 \cdot g \cdot m o l^{- 1}} = 1.05 \cdot m o l$ $"Moles of bromine"=(84.1*g)/(79.90*g*mol^-1)=1.05*mol$ .

And we divide these molar quantities by THE SMALLEST such quantity to get an empirical formula of $C H_{3} B r$ $CH_3Br$ . This is quite probably $methyl bromide$ $"methyl bromide"$ but we require an estimate of molecular mass before we address the molecular formula.

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A compound is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine. What is its empirical formula?

1 Answer

Explanation:

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