Question

0.105g of a given protein, containing unknown amounts of C H and O reacts with 02 to produce 0.257g of CO2 (44.01g/mol) and 0.035g of H2O (18.02g/mol), what is the empirical formula of the protein?

please explain


I just had this on an exam and had no clue how to do it, I need help

Answer 1

The answer to your question is C₃H₂O₁ or C₃H₂O

Explanation:

Data

mass of protein = 0.105 g

elements = C, H, O

mass of CO₂ = 0.257 g

molar mass of CO₂ = 44.01 g/mol

mass of water = 0.035 g

molar mass of water = 18.02 g/mol

empirical formula = ?

Process

1.- Calculate the mass and moles of carbon

44 g of CO₂ ------------ 12 g of C

0.257 g of CO₂------------- x

x = (0.257 x 12)/44

x = 3.084/44

x = 0.07 g of Carbon

12 g of C ----------------- 1 mol

0.07 g of C ------------- x

x = (0.07 x 1) / 12

x = 0.0058 moles of C

2.- Calculate the mass and moles of H

18 g of H₂O --------------- 2 g of H

0.035 g of H₂O ---------- x

x = (0.035 x 2) / 18

x = 0.0039 g of H₂

1 g of H ------------------- 1 mol of H

0.0039 g of H --------- x

x = 0.0039 moles of H

3.- Calculate the mass and moles of O₂

mass of O₂ = 0.105g - 0.07 - 0.0039g

= 0.0311 g of O₂

16 g of O₂ -------------- 1 mol

0.0311 f of O₂ --------- x

x = (0.0311 x 1)/16

x = 0.0019 moles of O₂

4.- Divide by the lowest number of moles

Carbon = 0.0058 / 0.0019 = 3.0

Hydrogen = 0.0039 / 0.0019 = 2.0

Oxigen = 0.0019 / 0.0019 = 1

5.- Write the empirical formula

C₃H₂O₁ or C₃H₂O