What is the molecular formula of a material with percentage composition $C : 40.5 %$ $C:40.5%$ , $H : 6.67 %$ $H:6.67%$ , $O : 53.33 %$ $O:53.33%$ , and molecular mass of $180.0 \cdot g \cdot m o l^{- 1}$ $180.0gmol^-1$ ?

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What is the molecular formula of a material with percentage composition $C : 40.5 %$ $C:40.5%$ , $H : 6.67 %$ $H:6.67%$ , $O : 53.33 %$ $O:53.33%$ , and molecular mass of $180.0 \cdot g \cdot m o l^{- 1}$ $180.0gmol^-1$ ?

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Answer 1 · 2017-05-07T15:57:20

This is a standard question, and as a routine we ASSUME a mass of $100 \cdot g$ $100*g$ of unknown compound.......and get a molecular formula of $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ .

Explanation:

So if there are $100 \cdot g$ $100*g$ of unknown compound, we can access the empirical formula given the percentage composition:

$Moles of carbon = \frac{40.0 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}} = 3.33 \cdot m o l$ $"Moles of carbon"=(40.0*g)/(12.011*g*mol^-1)=3.33*mol$

$Moles of hydrogen = \frac{6.67 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}} = 6.62 \cdot m o l$ $"Moles of hydrogen"=(6.67*g)/(1.00794*g*mol^-1)=6.62*mol$

$Moles of oxygen = \frac{53.33 \cdot g}{16.00 \cdot g \cdot m o l^{- 1}} = 3.33 \cdot m o l$ $"Moles of oxygen"=(53.33*g)/(16.00*g*mol^-1)=3.33*mol$

We divide thru by the LOWEST molar quantity, that of oxygen/carbon to give an empirical formula of:

$C H_{2} O$ $CH_2O$ , which represents the simplest whole number ratio defining constituent atoms in a species, i.e. the $empirical formula$ $"empirical formula"$ .

But we also have the molecular mass, $180 \cdot g \cdot m o l^{- 1}$ $180*g*mol^-1$ .

Now it is a fact that the molecular formula is always a whole number multiple of the empirical formula, and thus..............

$180 \cdot g \cdot m o l^{- 1} = n \times (12.011 + 2 \times 1.00794 + 15.999) \cdot g \cdot m o l^{- 1}$ $180*g*mol^-1=nxx(12.011+2xx1.00794+15.999)*g*mol^-1$ ,....................... and so we solve for $n$ $n$ . Clearly, $n = 6$ $n=6$ , and the molecular formula is $C_{6} H_{12} O_{6}$ $C_6H_12O_6$ .

Just on a practical issue, an analyst would normally NEVER give you the percentage oxygen by mass. Most of the time you get $% C, H, N$ $%C,H,N$ , and if their sum is not $100 %$ $100%$ , then the BALANCE is assumed to be due to $oxygen$ $"oxygen"$ .

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What is the molecular formula of a material with percentage composition $C : 40.5 %$ $C:40.5%$ , $H : 6.67 %$ $H:6.67%$ , $O : 53.33 %$ $O:53.33%$ , and molecular mass of $180.0 \cdot g \cdot m o l^{- 1}$ $180.0gmol^-1$ ?

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What is the molecular formula of a material with percentage composition C:40.5%C:40.5%, H:6.67%H:6.67%, O:53.33%O:53.33%, and molecular mass of 180.0⋅g⋅mol−1180.0*g*mol^-1?

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What is the molecular formula of a material with percentage composition $C : 40.5 %$ $C:40.5%$ , $H : 6.67 %$ $H:6.67%$ , $O : 53.33 %$ $O:53.33%$ , and molecular mass of $180.0 \cdot g \cdot m o l^{- 1}$ $180.0gmol^-1$ ?