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/mol. Add subscripts to complete the formulas.

Empirical formula: CHO :  

Molecular formula: CHO: 

Calculate the molar mass of CO₂.

12.0107 g/mol + 2(15.9994g/mol) = 44.0095 g/mol

Calculate the molar mass of H₂O.

2(1.0079 g/mol) + 15.9994 g/mol = 18.0152 g/mol

Convert the masses of CO₂ and H₂O to masses of C and H, respectively.

1. Begin with the mass of CO₂.

2. Use the molar mass of CO₂ as a conversion factor to convert grams of CO₂ to moles of CO₂.

3. Use a mole ratio to convert moles of CO₂ into moles of C.

4. Use the molar mass of C as a conversion factor to convert moles of C to grams of C.

m_C = (7.86 g CO₂) (1 mol /44.0095 g) ( 1 mol C / 1 mol CO₂) (12.0107 g/mol)

m_C = 2.1451 g

1. Begin with the mass of H₂O.

2. Use the molar mass of H₂O as a conversion factor to convert grams of H₂O to moles of H₂O.

3. Use a mole ratio to convert moles of H₂O into moles of H.

4. Use the molar mass of H as a conversion factor to convert moles of H to grams of H.

m_H = (2.14 g H₂O) ( 1 mol / 18.0152 g) ( 2 mol H/ 1 mol H₂O) (1.0079 g/mol)

m_H = 0.23945 g

Use the combustion equation in order to find the mass of oxygen. Note the m_tot represents the mass of organic compound.

m_tot = m_C + m_H + m_O

m_O = m_tot - m_C - m_H

m_O = 5.24 g - 2.1451 g - 0.23945 g

m_O = 2.85544 g

1. Begin with the mass of C.

2. Use the molar mass of C as a conversion factor to convert grams of C to moles of C.

moles C = (2.1451 g C) ( 1 mol C /12.0107 g C)

moles C = 0.17860 mol C

1. Begin with the mass of H.

2. Use the molar mass of H as a conversion factor to convert grams of H to moles of H.

moles H = (0.23945 g H) ( 1 mol H /1.0079 g H)

moles H = 0.23757 mol H

1. Begin with the mass of O.

2. Use the molar mass of O as a conversion factor to convert grams of O to moles of O.

moles O = (2.85544 g O) ( 1 mol O /15.9994 g O)

moles O = 0.17847 mol O

Determine the number of moles of each element relative to the elements. Divide all quantities by the smallest amout of moles.

(0.17860 mol / 0.17847 mol) C = 1 C

(0.23757 mol /0.17847 mol) H = 4/3H

(0.17847 mol /0.17847 mol) O = 1 O

Eliminate fractional coefficients by multiplying all coefficients by 3.

3(1) C = 3 C

3(4/3)H = 4 H

3(1) O = 3 O

Therefore, the empirical formula is C₃H₄O₃.

Calculate the molar mass of the empirical formula.

3(12.0107 g/mol) + 4(1.0079 g/mol) + 3(15.9994 g/mol) = 88.0619 g/mol

A molecular formula is always a whole number multiple of the empirical formula. Divide the molecular molar mass by the empirical molar mass to get this whole number.

n = 176 g/mol / 88.0619 g/mol

n = 2

Multiply the subscripts of the empirical formula by n in order to obtain the molecular formula.

2(3)C = 6C

2(4)H = 8H

2(3)O = 6O

Therefore, the molecular formula is C₆H₈O₆.