Chemistry, 14.04.2020 20:14 kalebbenton15

Ascorbic acid (vitamin C) contains C, H, and O. In one combustion analysis, 5.24 g of ascorbic acid yields 7.86 g CO2 and 2.14 g H2O. Calculate the empirical formula and molecular formula of ascorbic acid given that its molar mass is about 176 g.

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Answer from: lilisncz

Answer : The molecular formula for the given organic compound is C_6H_6O_6

Explanation :

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=7.86g

Mass of H_2O=2.14g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 7.86 g of carbon dioxide, $\frac{12}{44}\times 7.86=2.14g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 2.14 g of water, $\frac{2}{18}\times 2.14=0.238$ of hydrogen will be contained.

Mass of oxygen in the compound = (5.24) - (2.14 + 0.238) = 2.86 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.14g}{12g/mole}=0.178moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.238g}{1g/mole}=0.238moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{2.86g}{16g/mole}=0.178moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.178 moles.

For Carbon = $\frac{0.178}{0.178}=1$

For Hydrogen = $\frac{0.238}{0.178}=1.33\approx 1$

For Oxygen = $\frac{0.178}{0.178}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 1 : 1

The empirical formula for the given compound is C_1H_1O_1=CHO

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 176 g/mol

Mass of empirical formula = 29 g/mol

Putting values in above equation, we get:

$n=\frac{176}{29}=6$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 6)}H_{(1\times 6)}O_{(1\times 6)}=C_6H_6O_6$

Thus, the molecular formula for the given organic compound is C_6H_6O_6

Answer from: Quest

output?

explanation:

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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Ascorbic acid (vitamin C) contains C, H, and O. In one combustion analysis, 5.24 g of ascorbic acid...