Answer:
% composition of glucose = 85 %
Explanation:
Given Data
moles of glucose (C₆H₁₂O₆) = 1.3 mole
Total mass of mixture = 276 g
percent composition of glucose (C₆H₁₂O₆) = ?
Solution
Number of moles are given so we find out the mass of C₆H₁₂O₆
<em>mass = moles × molar mass</em>
Molar mass of glucose = 180.156 g/mol
mass = 1.3 mol × 180.156 g/mol
mass = 234.2 g
Now we find out the percent composition of glucose
% composition = (mass of compound ÷ mass of mixture) × 100
% composition = ( 234.2 g ÷ 276 g) × 100
% composition = 0.85 × 100
% composition = 85 %
