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Novosadov [1.4K]
3 years ago
13

A compound with the empirical formula ch2 has a molar mass of 98 g/mol what is the molecular formula for this compound ?

Chemistry
1 answer:
madam [21]3 years ago
7 0
Approximate molecular masses:
Molecular mass of C = 12
Molecular mass of H = 1

Let n = moles required for CH₂.
Then
nCH₂ = 98
n(12 + 2*1) = 98
14n = 98
n = 7

Answer: The molecular formula is 7CH₂

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2 years ago
Why did scientists in the 1800s decide that the list of known elements needed to be more
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8 0
3 years ago
Consider the following reaction: NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O + CO2 How many g of CO2 would be produced from the complete r
Alenkasestr [34]
<h2>1.25 g of CO_2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC_3H_3O_2 with excess NaHCO_3 </h2>

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}    

0.833M=\frac{\text{Moles of} HC_3H_3O_2\times 1000}{25ml}\\\\\text{Moles of} HC_3H_3O_2 =\frac{0.833mol/L\times 25}{1000}=0.0208mol

NaHCO_3+HC_2H_3O_2\rightarrow NaC_2H_3O_2+H_2O+CO_2

According to stoichiometry:

1 mole of HC_2H_3O_2 will give = 1 mole of CO_2

0.0208 moles of HC_2H_3O_2 will give =\frac{1}{1}\times 0.0208=0.0208 moles of CO_2

Mass of HC_2H_3O_2=moles\times {\text {molar mass}}=0.0208\times 60g/mol=1.25g

Thus 1.25 g of CO_2 would be produced from the complete reaction of 25 mL of 0.833 mol/L HC_3H_3O_2 with excess NaHCO_3

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