4. How many hydrogen atoms are in 345.3 g of iron(III) hydrogen phosphate? 5. How many atoms of chlorine are in 550.0 mL of carbon tetrachloride? The density of carbon tetrachloride is 1.59 g/mL.

Answer 1

4. The chemical formula for iron(iii) hydrogen phosphate is Fe₂(HPO₄)_3. It has a molar mass of 399.62g/mol. So there are 345.2g/399.62g/mol = 0.8641 moles of Fe₂(HPO₄)₃. Each molecule of Fe₂(HPO₄)₃ has 3 hydrogen atoms, so there must be 0.8641 * 3 = 2.592 moles of hydrogen atoms. Multiply by Avogadro's number to find the exact number of hydrogen atoms.

5. You have 550.0mL * 1.59g/mL = 874.5g of carbon tetrachloride. Similar to the problem above, divide by the molar mass to find moles of carbon tetrachloride. Carbon tetrachloride has 4 Cl, so multiply moles of carbon tetrachloride by 4. Multiply by Avogadro's number to find the exact number of chlorine atoms.

Answer 2

4) The chemical formula for iron(III) hydrogen phosphate is Fe₂(HPO₄)₃. We can calculate its molar mass:

MM_Fe2(HPO4)3 = 2(MM_iron) + 3(MM_hydrogen) + 3(MM_phosphorus) + 12(MM_oxygen)

= 2(55.84g/mol) + 3(1.008g/mol) + 3(30.97g/mol) + 12(16.00g/mol) = 399.614g/mol

To convert from mass to moles, we divide by the molar mass:

mol_Fe2(HPO4)3 = g_Fe2(HPO4)3/MM_Fe2(HPO4)3 = (345.3g)/(399.614g/mol) = 0.8641mol_Fe2(HPO4)3

We know each molecule of Fe₂(HPO₄)₃ has 3 hydrogen atoms, so we can find the moles of hydrogen:

mol_hydrogen = mol_Fe2(HPO4)3 * (mol_hydrogen/mol_Fe2(HPO4)3)

= 0.8641mol_Fe2(HPO4)3 * (3mol_H/1mol_Fe2(HPO4)3) = 2.592mol_hydrogen

To get the number of atoms, we multiply by Avogadro's number:

molecules_hydrogen = (2.592mol_hydrogen)(6.022x10²³molecules/mol) = 1.561x10²⁴ molecules_hydrogen

5) The chemical formula for carbon tetrachloride is CCl₄. We can calculate its molar mass:

MM_CCl4 = (MM_carbon) + 4(MM_chlorine) = (12.01g/mol) + 4(35.45g/mol) = 153.81g/mol

To find the number of moles, we first find the mass of CCl₄:

g_CCl4 = volume_CCl4 * density_CCl4 = (550.0mL)(1.59g/mL) = 874.5g_CCl4

Like in (4), we now divide by the molar mass:

mol_CCl4 = g_CCl4/MM_CCl4 = (874.5g)/(153.81g/mol) = 5.686mol_CCl4

We know each molecule of CCl₄ has 4 chlorine atoms, so we can find the moles of chlorine:

mol_chlorine = mol_CCl4 * (mol_hydrogen/mol_Fe2(HPO4)3)

= 5.686mol_CCl4 * (4mol_Cl/1mol_CCl4) = 22.74mol_chlorine

To get the number of atoms, we multiply by Avogadro's number:

molecules_chlorine = (22.74mol_chlorine)(6.022x10²³molecules/mol) = 1.370x10²⁵ molecules_chlorine