Question

please help!!How many atoms of carbon are present in 34.5 g of caffeine, C3H10N4O2?
8.57X10^23
O 2.68X10^25
O 4.83X10^23
O 1.08X10^24

Answer 1

Answer:

• Molar mass of caffeine:

$= { \rm{(12 \times 3) + (1\times 60) + (14 \times 4) + (16 \times 2)}} \\ \\ = { \rm{194 \: g}}$

• From avogadro's number:

${ \rm{134 \: g \: \dashrightarrow \: 6.02 \times {10}^{23} \: atoms}} \\ { \rm{34.5 \: g \: \dashrightarrow \: ( \frac{34.5}{194} \times 6.02 \times {10}^{23}) \: atoms }} \\ \\{ \boxed { \rm{ = 1.08 \times {10}^{23} \: atoms}}}$