Question #e2fd1

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Question #e2fd1

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Answer 1 · 2017-06-19T15:08:28

$C_{7} H_{7} N O$ $C_7H_7NO$

Explanation:

As with all of these problems, we assume a $100 \cdot g$ $100*g$ mass of unknown compound, and we interrogate its atomic composition...

$Moles of carbon = \frac{69.40 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}} = 5.78 \cdot m o l$ $"Moles of carbon"=(69.40*g)/(12.011*g*mol^-1)=5.78*mol$ .

$Moles of hydrogen = \frac{5.825 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}} = 5.78 \cdot m o l$ $"Moles of hydrogen"=(5.825*g)/(1.00794*g*mol^-1)=5.78*mol$ .

$Moles of oxygen = \frac{13.21 \cdot g}{15.999 \cdot g \cdot m o l^{- 1}} = 0.826 \cdot m o l$ $"Moles of oxygen"=(13.21*g)/(15.999*g*mol^-1)=0.826*mol$ .

$Moles of nitrogen = \frac{11.57 \cdot g}{14.01 \cdot g \cdot m o l^{- 1}} = 0.826 \cdot m o l$ $"Moles of nitrogen"=(11.57*g)/(14.01*g*mol^-1)=0.826*mol$ .

And if we divide thru by the SMALLEST molar quantity, i.e. $0.826 \cdot m o l$ $0.826*mol$ to get a trial empirical formula of......

$C_{7} H_{7} N O$ $C_7H_7NO$ ............

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Question #e2fd1

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