Question

now its your turn... • environment / air pollution carbon dioxide (co₂) a car engine produces 88.0 g of co₂ during a short trip. • how many molecules of co₂ were released? • how many oxygen atoms were in that emission?

Explanation:

Sub - Question 1: How many molecules of \(CO_2\) were released?

Step 1: Calculate molar mass of \(CO_2\)

Molar mass of \(C\) is \(12.01\space g/mol\), molar mass of \(O\) is \(16.00\space g/mol\). For \(CO_2\), molar mass \(M = 12.01+ 2\times16.00=44.01\space g/mol\)

Step 2: Calculate moles of \(CO_2\)

Given mass \(m = 88.0\space g\), moles \(n=\frac{m}{M}\). Substituting values, \(n=\frac{88.0\space g}{44.01\space g/mol}\approx2.00\space mol\)

Step 3: Calculate number of molecules

Using Avogadro's number \(N_A = 6.022\times 10^{23}\space molecules/mol\), number of molecules \(N=n\times N_A\). Substituting \(n = 2.00\space mol\), \(N = 2.00\space mol\times6.022\times 10^{23}\space molecules/mol = 1.2044\times 10^{24}\space molecules\) (approx \(1.20\times 10^{24}\) or more precisely \(1.204\times 10^{24}\))

Sub - Question 2: How many oxygen atoms were in that emission?

Step 1: Recall the formula of \(CO_2\)

In one molecule of \(CO_2\), there are 2 oxygen atoms.

Step 2: Use number of \(CO_2\) molecules

From part 1, number of \(CO_2\) molecules \(N = 1.2044\times 10^{24}\). Number of oxygen atoms \(N_O=2\times N\)

Step 3: Calculate number of oxygen atoms

\(N_O = 2\times1.2044\times 10^{24}=2.4088\times 10^{24}\space atoms\) (approx \(2.41\times 10^{24}\))

Answer:

(Sub - Question 1):
The number of \(CO_2\) molecules released is approximately \(\boldsymbol{1.20\times 10^{24}}\) (or more precisely \(1.204\times 10^{24}\)) molecules.