Answer:
[tex]C_{18}H_{24}O_3[/tex]
Explanation:
Hello,
In this case, combustion analyses help us to determine the empirical formula of a compound via the quantification of the released carbon dioxide and water since the law of conservation of mass is leveraged to attain it. In such a way, as 36.86 g of carbon dioxide were obtained, this directly represents the mass of carbon present in the sample, thus, we first compute the moles of carbon:
[tex]n_{C}=36.86gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.838molC[/tex]
Then, into the water one could find the moles of hydrogen:
[tex]n_{H_2O}=10.06gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} =1.12molH[/tex]
Now, we compute the moles of oxygen by firstly computing the mass of oxygen:
[tex]m_{O}=13.42g-36.86gCO_2*\frac{12gC}{44gCO_2} -10.06 gH_2O*\frac{2gH}{18gH_2O} =2.25gO[/tex]
[tex]n_O=2.25gO*\frac{1molO}{16gO} =0.141molO[/tex]
Then, we have the mole ratio:
[tex]C_{0.838}H_{1.12}O_{0.141}\rightarrow C_6H_8O[/tex]
Whose molar mass is 12x6+1x8+16=96 g/mol, but the whole compound molar mass is 288.38, the factor is 288.38/96 =3, Therefore the formula:
[tex]C_{18}H_{24}O_3[/tex]
Regards.
Answer:
Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]
Explanation:
[tex](\frac{36.86gCO_{2} }{44g/molCO_{2} })[/tex] [tex](\frac{10.06gH_{2}O}{18g/molH_{2}O} )[/tex]
(0.8377mol of CO2) (0.556 mol of H2O)
mole ratio of CO2 to H2O = 1.5 : 1 ≅ 3 : 2
[tex](CO_{2} )_{3} , (H_{2} O)_{2}[/tex]
⇒ [tex](C_{3} H_{4} O_{8} )_{x}[/tex] = 288.38
(12 × 3 + 1 × 4 + 16 × 8)x = 288.38
168x = 288.38
x = 1.71 ≅ 2
∴ Molecular mass = [tex]C_{6} H_{8} O_{16}[/tex]
