How many molecules are in each sample?

(a) 3.5 g H2O

(b) 56.1 g N2

(c) 89 g CCI4

(d) 19 g C6H2O6

Solution 53P

Here, we have to calculate the number of molecules in each sample.

3.5 g H2OStep 1: Calculating the molar mass of of H2O

molar mass = 1 x (16.00) + 2x (1.008)

= 16 + 2.016 = 18.016 g/mol

Step 2: Calculating the number of molecules in 3.5 g of H2O

1.0 mol of H2O = 6.02 x 1023 molecules of H2O

1.0 mol of H2O = 18.016 g

The conversion factors are :

, (6.02 x1023 molecules of of H2O /1.0 mol of of H2O)

Thus, 3.5 g of of H2O contains

= 3.5 g x () x (6.02 x1023 molecules of of H2O /1.0 mol of of H2O)

= (21.077/18.016) x 1023 molecules of of H2O

= 1.169 x 1023 molecules of of H2O

Therefore, 3.5 g of of H2O contains 1.169 x 1023 molecules of H2O.

(b) 56.1 g N2

Step 1: Calculating the molar mass of N2

molar mass = 2 x (14.006)

= 16 + 2.016 = 28.01 g/mol

Step 2: Calculating the number of molecules in 56.1 g N2

1.0 mol of N2 = 6.02 x 1023 molecules N2

1.0 mol of N2 = 28.01 g

The conversion factors are :

, (6.02 x1023 molecules of N2 /1.0 mol of N2)

Thus, 56.1 g of N2 contains

= 56.1 g x () x (6.02 x1023 molecules of N2 /1.0 mol of N2)

= (337.834/28.01) x 1023 molecules of N2

= 12.06 x 1023 molecules of N2

Therefore, 56.1 g of N2 contains 12.06 x 1023 molecules of N2.

(c) 89 g CCI4

Step 1: Calculating the molar mass of CCl4

molar mass = 1x (12.01) + 4 x(35.45)

= 153.81 g/mol

Step 2: calculating the number of molecules in 89 g CCI4

1.0 mol of CCI4 = 6.02 x 1023 molecules CCI4

1.0 mol of CCI4 = 153.81 g

The conversion factors are :

, (6.02 x1023 molecules of CCI4 /1.0 mol of CCI4)

Thus, 89 g CCI4 contains

= 89 g x () x (6.02 x1023 molecules of CCI4 /1.0 mol of CCI4)

= (535.958/153.81) x 1023 molecules of CCl4

= 3.484 x 1023 molecules of CCl4

Therefore, 89g of CCl4 contains 3.484 x 1023 molecules of CCl4.

(d) 19 g C6H2O6

Step 1: Calculating the molar mass of C6H2O6

molar mass = 6x (12.01) +2 x(1.008)+6 x(16.0)

= 170.076 g/mol