How many molecules of CO₂ do you have in 28.0 g of CO₂?

Question

| General Chemistry 2

Accepted Answer

n_CO2 = $\frac{m_{CO 2}}{M_{CO 2}}$ and n_CO2 = $\frac{N_{CO 2}}{N_{A}}$

So,

\frac{N_{CO 2}}{N_{A}}

=

\frac{m_{CO 2}}{M_{CO 2}}

N_CO2 =

\frac{m_{CO 2}}{M_{CO 2}}

x N_A =

\frac{m_{CO 2}}{M_{C} + 2 M_{O}}

x N_A =

\frac{28.0}{44.0}

x 6.022 x 10²³ = 3.83 x 10²³ atoms

Mass-mole-number relationship - 1 | Chemical Calculations