Question

question 18 the chemical formula for glucose is c₆h₁₂o₆. what is the percent composition for oxygen in glucose? molar masses to use: c = 12.011; h = 1.008; o = 15.999 a 180.156% b 43.28% c 53.28% d 6.71% what is the percent composition of c₂h₂? (molar masses to use: c=12.011 and h=1.008) a c = 92.26% b c = 12.011% c c = 24.02% d h = 1.008% e h = 7.74% f h = 2.01%

Explanation:

Step1: Calculate molar mass of glucose

Molar mass of \( C_6H_{12}O_6 \) = \( 6\times C + 12\times H + 6\times O \)
\( = 6\times12.011 + 12\times1.008 + 6\times15.999 \)
\( = 72.066 + 12.096 + 95.994 \)
\( = 180.156 \)

Step2: Calculate mass of oxygen in glucose

Mass of O = \( 6\times15.999 = 95.994 \)

Step3: Calculate percent composition of O

Percent composition = \( \frac{\text{Mass of O}}{\text{Molar mass of glucose}} \times 100\% \)
\( = \frac{95.994}{180.156} \times 100\% \approx 53.28\% \)

Answer:

C (53.28%)