How many molecules of CO 2 do you have in 28.0 g of CO 2 ? | General Chemistry 2

n CO2 = m CO 2 M CO 2 and n CO2 = N CO 2 N A So, N CO 2 N A = m CO 2 M CO 2 N CO2 = m CO 2 M CO 2 x N A = m CO 2 M C + 2 M O x N A = 28 . 0 44 . 0 x 6.022 x 10 23 = 3.83 x 10 23 atoms

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Mass-mole-number relationship - 1 | Chemical Calculations

Chapter 2 - Chemical Calculations

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

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Solution

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

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Mass-mole-number relationship - 1 | Chemical Calculations

General Chemistry 2 - Mass-mole-number relationship - 1