Respuesta :
Answer:
The molecular formula of X is given as [tex]C_7 H_6 O_3[/tex]
Explanation:
[tex]Moles $C O_{2}=\frac{\text { mass }}{\text { molar mass }}=\frac{13.39 \mathrm{g}}{44.01 \mathrm{g} \text { per mole }}=0.304 \mathrm{mol}$\\\\moles $\mathrm{C}=$ moles $\mathrm{CO}_{2}=0.304 \mathrm{mol}$[/tex]
[tex]mass $C=$ moles $\times$ molar mass $=0.304 \mathrm{mol} \times 12 \frac{g}{m o l}=3.65g$\\\\moles $\mathrm{H}_{2} \mathrm{O}=\frac{2.35 \mathrm{g}}{18.02 \mathrm{g} \text { permole }}=0.130 \mathrm{mol}$\\\\moles $\mathrm{H}=2 \times$ moles $\mathrm{H}_{2} \mathrm{O}=0.130 \times 2=0.260 \mathrm{mol}$\\\\Mass $\mathrm{H}=0.260 \mathrm{mol} \times 1.008 \frac{g}{\mathrm{mol}}=0.262 \mathrm{g}$[/tex]
mass O = Total mass of the compound - (mass of C + mass of H)
=6.0 g - ( 3.65 + 0.262 ) g
=2.09 g
[tex]moles $O=\frac{2.09 g}{16 g \text { per mole }}=0.131 \mathrm{mol}$[/tex]
Least moles is for O that is 0.131mol and dividing all by the least we get
[tex]$\begin{aligned} C &=\frac{0.304}{0.131}=2.3 \\\\ H &=\frac{0.260}{0.131}=2 \\\\ O &=\frac{0.131}{0.131}=1 \end{aligned}$[/tex]
Since 2.3 is a fraction it has to be converted to a whole number so we multiply all the answers by 3
[tex]\\$C 2.3 \times 3=7$\\\\$H 2 \times 3=6$\\\\$O 1 \times 3=3$[/tex]
So the empirical formula is [tex]C_7 H_6 O_3[/tex]
Empirical formula mass
[tex]=(7 \times 12) +(6\times1.008)+(3\times16)=138.048g[/tex]
[tex]$n=\frac{\text { molar mass }}{\text { empirical formula mass }}=\frac{138}{138.048}=1$[/tex]
Molecular formula =n × empirical formula
[tex]=1 \times C_7 H_6 O_3[/tex]
Compound X = [tex]C_7 H_6 O_3[/tex] is the Answer
The compound X that have a molecular molar mass of 138 g/mol is [tex]\rm \bold{C_7H_6O_3}[/tex]
Given here,
Compound x - 6g
malar mass of x - 138g/mol
Carbon dioxide - 13.39 g
water - 2.35g
So,
The moles of Water will be 0.13 mol
The moles of Hydrogen will be 0.26
Mass of hydrogen = 0.26 g
The moles of carbon dioxide will be 0.304
The moles of carbon will be 0.304
Mass of Carbon = 3.65 g
Mass of Oxygen = Total mass of the compound - (mass of C + mass of H)
= 6.0 g - ( 3.65 + 0.262 ) g
= 2.09 g
Moles of oxygen = 0.131
Since Oxygen has least moles,
To find the molar ratio of constituent atom, divide moles of atom from oxygen.
[tex]\rm \bold { C = \frac{0.304}{0.131} } = \bold {2.3 }\\\\\rm \bold { C = \frac{0.26 }{0.131} } = \bold {2}\\\\\rm \bold { C = \frac{0.131}{0.131} } = \bold {1}[/tex]
convert 2.3 into whole number
[tex]\rm \bold { C \Rightarrow 2.3 \times 3 = 7}\\\rm \bold { H \Rightarrow 2 \times 3 = 6}\\\rm \bold { O \Rightarrow 1 \times 3 = 3}[/tex]
Hence formula will be [tex]\rm \bold{C_7H_6O_3}[/tex]
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