Question

question 13 of 26
determine the mass in grams of 1.6 × 10⁻³ mol glucose, c₆h₁₂o₆
starting amount
add factor answer reset
84.18 0.29 mol c₆h₁₂o₆
6.022 × 10²³ g c₆h₁₂o₆/mol
1 g c₆h₁₂o₆
180.18 molecules c₆h₁₂o₆
1.6 × 10⁻³
0.13
9.6 × 10²⁰
12.01

Explanation:

Step1: Identify starting amount

Starting amount: $1.6 \times 10^{-3}\ \text{mol}\ \text{C}_6\text{H}_{12}\text{O}_6$

Step2: Find molar mass of glucose

Molar mass of $\text{C}_6\text{H}_{12}\text{O}_6$:
$6(12.01) + 12(1.008) + 6(16.00) = 180.18\ \text{g/mol}$

Step3: Calculate mass via conversion

Mass = Moles × Molar Mass
$1.6 \times 10^{-3}\ \text{mol} \times 180.18\ \frac{\text{g}\ \text{C}_6\text{H}_{12}\text{O}_6}{\text{mol}\ \text{C}_6\text{H}_{12}\text{O}_6}$

Answer:

$0.29\ \text{g}\ \text{C}_6\text{H}_{12}\text{O}_6$

(Note: Filled into the structure:
Starting amount box: $1.6 \times 10^{-3}\ \text{mol}\ \text{C}_6\text{H}_{12}\text{O}_6$
Fraction numerator: $180.18\ \text{g}\ \text{C}_6\text{H}_{12}\text{O}_6$
Fraction denominator: $1\ \text{mol}\ \text{C}_6\text{H}_{12}\text{O}_6$
Final calculated answer: $0.29\ \text{g}\ \text{C}_6\text{H}_{12}\text{O}_6$)