Question #39025

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Question #39025

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Answer 1 · 2017-08-19T07:26:02

$C H_{4}$ $CH_4$

Explanation:

As always, it is useful to assume $100 \cdot g$ $100*g$ of unknown compound, and then we interrogate its atomic composition:

$Moles of C$ $"Moles of C"$ $=$ $=$ $\frac{75 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}} = 6.24 \cdot m o l \cdot C$ $(75*g)/(12.011*g*mol^-1)=6.24*mol*C$

$Moles of H$ $"Moles of H"$ $=$ $=$ $\frac{25 \cdot g}{100794 \cdot g \cdot m o l^{- 1}} = 24.8 \cdot m o l \cdot H$ $(25*g)/(100794*g*mol^-1)=24.8*mol*H$

And we divide thru by the SMALLEST molar quantity to get......

$C_{\frac{6.24 \cdot m o l}{6.24 \cdot m o l}} H_{\frac{24.8 \cdot m o l}{6.24 \cdot m o l}} = C H_{4}$ $C_((6.24*mol)/(6.24*mol))H_((24.8*mol)/(6.24*mol))=CH_4$ .

Clearly, the question was devised purely for pedagogy, as the unknown compound is methane on the basis of these data - it would be hard to collect combustion data on a gas.

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Question #39025

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