Answer
C: $ 3.01 \times 10^{22} \space atoms $
H: $ 6.02 \times 10^{22} \space atoms $
O: $ 3.01 \times 10^{22} \space atoms $
Work Step by Step
1. Find the amount of moles of glucose:
- Calculate or find the molar mass for $ C_6H_{12}O_6 $:
$ C_6H_{12}O_6 $ : ( 1.008 $\times$ 12 )+ ( 12.01 $\times$ 6 )+ ( 16.00 $\times$ 6 )= 180.16 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 1.50 \space g \times \frac{1 \space mole}{ 180.16 \space g} = 8.33 \times 10^{-3} \space mole$$
2. Calculate the amount of atoms of each element:
- Each $ C_6H_{12}O_6 $ has 6 C atoms, thus:$$ 8.33 \times 10^{-3} \space mole \space C_6H_{12}O_6 \times \frac{ 6 \space moles \ C }{1 \space mole \space C_6H_{12}O_6 } \times \frac{6.022 \times 10^{23} \space atoms }{1 \space mole} = 3.01 \times 10^{22} \space atoms $$
- Each $ C_6H_{12}O_6 $ has 12 H atoms, thus:$$ 8.33 \times 10^{-3} \space mole \space C_6H_{12}O_6 \times \frac{ 12 \space moles \ H }{1 \space mole \space C_6H_{12}O_6 } \times \frac{6.02 \times 10^{23} \space atoms }{1 \space mole} = 6.02 \times 10^{22} \space atoms $$
- Each $ C_6H_{12}O_6 $ has 6 O atoms, thus:$$ 8.33 \times 10^{-3} \space mole \space C_6H_{12}O_6 \times \frac{ 6 \space moles \ O }{1 \space mole \space C_6H_{12}O_6 }\times \frac{6.022 \times 10^{23} \space atoms }{1 \space mole} = 3.01 \times 10^{22} \space atoms $$
