Question

now it’s your turn... • nutrition / food science • glucose (c₆h₁₂o₆) a small candy bar contains 12.0 g of glucose. • how many molecules of glucose does this represent? • how many hydrogen atoms are in that candy bar?

Explanation:

First Sub - Question: How many molecules of glucose does this represent?

Step 1: Calculate molar mass of glucose ($\ce{C6H12O6}$)

Molar mass of C: $12.01\ g/mol$, H: $1.008\ g/mol$, O: $16.00\ g/mol$.
Molar mass of $\ce{C6H12O6}$ = $6\times12.01 + 12\times1.008+6\times16.00$
= $72.06+12.096 + 96.00$ = $180.156\ g/mol$

Step 2: Calculate moles of glucose

Moles ($n$) = $\frac{mass}{molar\ mass}$ = $\frac{12.0\ g}{180.156\ g/mol}$ ≈ $0.0666\ mol$

Step 3: Calculate number of molecules

Using Avogadro's number ($N_A = 6.022\times10^{23}\ molecules/mol$)
Number of molecules ($N$) = $n\times N_A$ = $0.0666\ mol\times6.022\times10^{23}\ molecules/mol$ ≈ $4.01\times10^{22}\ molecules$

Step 1: From previous step, moles of glucose ($n_{glucose}$) ≈ $0.0666\ mol$

Step 2: Determine moles of H atoms

In 1 mole of $\ce{C6H12O6}$, there are 12 moles of H atoms. So moles of H ($n_H$) = $12\times n_{glucose}$ = $12\times0.0666\ mol$ = $0.7992\ mol$

Step 3: Calculate number of H atoms

Using Avogadro's number ($N_A = 6.022\times10^{23}\ atoms/mol$)
Number of H atoms ($N_H$) = $n_H\times N_A$ = $0.7992\ mol\times6.022\times10^{23}\ atoms/mol$ ≈ $4.81\times10^{23}\ atoms$

Answer:

Approximately $4.01\times10^{22}$ molecules of glucose.

Second Sub - Question: How many hydrogen atoms are in that candy bar?