Determine the number of molecules in 1.512 g of ammonium sulfate, (NH_4)_2 SO_4.

Question

Determine the number of molecules in 1.512 g of ammonium sulfate, (NH_4)_2 SO_4. Show setup with correct conversion factors: Calculate the number of molecules in 2.37 moles of CO_2.(Sec Section 9.5, examples 9.6) Show setup with correct conversion factor Calculate number atoms of Cu in 3.23 g of metal Cu (See Section 9.1O, examples 9.13). Show setup with correct conversion factors: Calculate number of oxygen atoms present in 3.123 g of CO_2. Show all steps by using conversion factors as done in the class by using mass of H_2O. (example 9.15) You calculate molar mass if needed.

Explanation / Answer

We calculate this with molar mass of ammonium sulfate to convert into moles, and then with moles of ammonium sulfate we calculate number of molecules with Avogadro’s number.

Grams of ammonium sulfate and molar mass ---> moles ----> number of molecules.

Molar mass of ammonium sulfate, we calculate this with molar mass found in periodic table and multiplying by number of each atom present in the molecule: 132.14 g/mol

Avogadro’s number: 6.022 x 1023 atoms or molecules. 1 mole of any element contains 6.022 x 1023 atoms and 1 mole of any molecular compound contains 6.022 x 1023 molecules.

1.512 g (NH4)2SO4 (1 mol / 132.14 g (NH4)2SO4) = 0.011 moles of (NH4)2SO4

If 1 mole of (NH4)2SO4 contains 6.022 x 1023 molecules, how many molecules will have 0.011 moles of (NH4)2SO4?

0.011 moles of (NH4)2SO4 (6.022 x 1023 molecules / 1 mole of (NH4)2SO4) = 6.62 x 1021 molecules

Number of molecules = 6.62 x 1021 molecules

Similar to the problem above, we calculate number of molecules from number of moles of carbon dioxide.

Moles of carbon dioxide ----> number of molecules in carbon dioxide (with Avogadro’s number)

If 1 mole of carbon dioxide contains 6.022 x 1023 molecules, how many molecules are present in 2.37 moles of CO2?

2.37 moles of CO2 (6.022 x 1023 molecules / 1 mole of CO2 = 1.427 x 1024 molecules

Number of molecules = 1.427 x 1024 molecules

Grams of copper and molar mass ---> number of moles ----> number of molecules.

Molar mass of copper: 63.546 g/mol

Avogadro’s number: 6.022 x 1023 atoms

3.23 g of Cu (1 mol / 63.546 g Cu) = 0.0508 moles of Cu.

0.0508 moles of Cu (6.022 x 1023 atoms / 1 mole of Cu) = 3.059 x 1022 atoms.

Number of atoms present in 3.23 g of Cu = 3.059 x 1022 atoms.

Grams of carbon dioxide and molar mass ---> number of moles of carbon dioxide ----> number of molecules of carbon dioxide ----> number of atoms of oxygen present in the sample.

Molar mass of carbon dioxide: 44.01 g/mol

Avogadro’s number: 6.022 x 1023 atoms/molecules

3.123 g of CO2 (1 mol / 44.01 g of CO2) = 0.071 moles of CO2

0.071 moles of CO2 (6.022 x 1023 atoms / 1 mol CO2) = 4.276 x 1022 molecules of CO2

Since 1 molecule of CO2 contains 2 atoms of oxygen, we multiply the number of molecules calculated before to obtain number of atoms of oxygen present in the carbon dioxide molecule.

2 x 4.276 x 1022 molecules = 8.55 x 1022 atoms of oxygen.

Number of atoms of oxygen= 8.55 x 1022 atoms

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