Answer
$1.10 \times 10^{26}$ atoms of Carbon.
$3.30 \times 10^{26}$ atoms of Hydrogen.
$5.50 \times 10^{25}$ atoms of Sulfur.
$5.50 \times 10^{25}$ atoms of Oxygen.
Work Step by Step
- Calculate or find the molar mass for $ (CH_3)_2SO $:
$ (CH_3)_2SO $ : ( 1.008 $\times$ 6 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 1 )+ ( 32.07 $\times$ 1 )= 78.14 g/mol
- Using the molar mass as a conversion factor, find the amount in moles:
$$ 7.14 \times 10^{3} \space g \times \frac{1 \space mole}{ 78.14 \space g} = 91.4 \space moles$$
- Each $ (CH_3)_2SO $ has 2 C atoms, thus:$$ 91.4 \space mole \space (CH_3)_2SO \times \frac{ 2 \space moles \ C }{1 \space mole \space (CH_3)_2SO } \times \frac{6.02 \times 10^{23} \space atoms }{1 \space mole} = 1.10 \times 10^{26} \space atoms $$
- Each $ (CH_3)_2SO $ has 6 H atoms, thus:$$ 91.4 \space mole \space (CH_3)_2SO \times \frac{ 6 \space moles \ H }{1 \space mole \space (CH_3)_2SO } \times \frac{6.022 \times 10^{23} \space atoms \ H }{1 \space mole \space H } = 3.30 \times 10^{26} \space moles \space H $$
- Each $ (CH_3)_2SO $ has 1 S atoms, thus:$$ 91.4 \space mole \space (CH_3)_2SO \times \frac{ 1 \space moles \ S }{1 \space mole \space (CH_3)_2SO } \times \frac{6.022 \times 10^{23} \space atoms \ S }{1 \space mole \space S } = 5.50 \times 10^{25} \space moles \space S $$
- Each $ (CH_3)_2SO $ has 1 O atoms, thus:$$ 91.4 \space mole \space (CH_3)_2SO \times \frac{ 1 \space moles \ O }{1 \space mole \space (CH_3)_2SO } \times \frac{6.022 \times 10^{23} \space atoms \ O }{1 \space mole \space O } = 5.50 \times 10^{25} \space moles \space O $$
