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8_murik_8 [283]
3 years ago
14

An unknown compound containing only C and H was burnt, yielding 10.2 g of CO2 and 6.3 g of H2O. With a molecular weight of about

30, what is the molecular formula?
Chemistry
1 answer:
Lisa [10]3 years ago
3 0

Answer:

C_2H_6

Explanation:

Hello.

In this case, we can see that the mass of carbon of the unknown compound comes from the yielded mass of carbon dioxide, thus, we compute the moles of carbon as follows:

m_C=10.2gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1mol C}{1molCO_2}=0.232 molC

Moreover, the mass of hydrogen comes from the yielded water, therefore we can also compute the moles of water:

m_H=6.3gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH_2}{1molH_2O}  =0.7molH

Then, to find the subscripts in the empirical formula, we divide by the moles of carbon as the smallest:

C:\frac{0.232}{0.232}=1\\ \\H:\frac{0.7}{0.232}=3

Whose molar mass is:

M_{CH_3}=12+1*3=15g/mol

Thus, the ratio of the molecular formula to the empirical formula is:

\frac{30}{15}=2

Therefore, the molecular formula is twice the empirical formula:

C_2H_6

Which is actually ethane.

Regards.

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Alex787 [66]

Answer:

M_{HCl}=7.96M

Explanation:

Hello,

In this case, since the neutralization reaction between HCl and NaOH is:

HCl+NaOH\rightarrow NaCl+H_2O

We notice a 1:1 molar ratio, for that reason, at the equivalence point we find:

n_{HCl}=n_{NaOH}

Thus in terms in molarities one could compute the concentration of HCl in the old bottle for the used NaOH for the neutralization as:

M_{HCl}V_{HCl}=M_{NaOH}V_{NaOH}\\\\M_{HCl}=\frac{M_{NaOH}V_{NaOH}}{V_{HCl}} =\frac{12.0M*9.95mL}{15mL}\\ \\M_{HCl}=7.96M

This value is lower than 37% HCl that in molarity is about 12 M, such difference is due to its high volatility.

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