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Question1. The following data was obtained when magnesium oxide was produced from the following reaction 2Mg+O2→2MgO2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{MgO}
Table 1: Masses of reactants and products \begin{tabular}{|l|l|} \hline Mass of Test Tube & 22.42 g \\ \hline Mass of Test Tube and Magnesium & 22.52 g \\ \hline Mass of Test Tube and Product & 23.01 g \\ \hline \end{tabular} a) Calculate the mass of magnesium reacted. 0.19 b) Calculate the mass of magnesium oxide produced. 0.59 g c) Calculate the mass of oxygen gas reacted. 0.49 g d) Calculate the experimental percent composition of magnesium and oxygen in magnesium oxide. 16.9%−%Mg;83.1%0=7.016.9 \%-\% \mathrm{Mg} ; 83.1 \%_{0}=7.0 e) What is the theoretical percent of magnesium in magnesium (HINT: this is the value for the percent composition using the %Mg=60.3%;20:39.7%\% \mathrm{Mg}=60.3 \% ; 20: 39.7 \%.
2. Calculate the molar mass of eqch of the following: a) Zinc chloride 136.28 g/mol136.28 \mathrm{~g} / \mathrm{mol} b) Sodium phosphate 163.94 g/mol163.94 \mathrm{~g} / \mathrm{mol}
3. What is the mass of 2.36 moles of silver nitrate? 401 g401 \mathrm{~g}
4. How many moles are in 5.62×10275.62 \times 10^{27} molecules of sodium chloride? 9.34×1039.34 \times 10^{3} moles.
5. How many molecules are in a 3.54 g sample of propane (C3H8)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) ? How many atoms are in this sample? 4.83×10224.83 \times 10^{22}
6. An inorganic salt is composed of 38.8%38.8 \% calcium, 20.0%20.0 \% phosphorus, and 41.2%41.2 \% oxygen. What is the empirical formula for this salt? Ca3P2O8\mathrm{Ca}_{3} \mathrm{P}_{2} \mathrm{O}_{8}
7. The empirical formula of a certain chemical is C6H2OCl2\mathrm{C}_{6} \mathrm{H}_{2} \mathrm{OCl}_{2}. If the molar mass is 322 g/mol322 \mathrm{~g} / \mathrm{mol}, what is its molecular formula? C12H4O2Cl4\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{O}_{2} \mathrm{Cl}_{4}

Studdy Solution

STEP 1

1. The chemical reaction is 2Mg+O2→2MgO2 \mathrm{Mg} + \mathrm{O}_{2} \rightarrow 2 \mathrm{MgO}.
2. The mass of the test tube is 22.42 g22.42 \, \text{g}.
3. The mass of the test tube and magnesium is 22.52 g22.52 \, \text{g}.
4. The mass of the test tube and product is 23.01 g23.01 \, \text{g}.
5. The molar masses of elements and compounds are known for calculations.
6. Avogadro's number is 6.022×1023 molecules/mol6.022 \times 10^{23} \, \text{molecules/mol}.

STEP 2

1. Calculate the mass of magnesium reacted.
2. Calculate the mass of magnesium oxide produced.
3. Calculate the mass of oxygen gas reacted.
4. Calculate the experimental percent composition of magnesium and oxygen in magnesium oxide.
5. Calculate the theoretical percent composition of magnesium in magnesium oxide.
6. Calculate the molar mass of given compounds.
7. Calculate the mass of a given number of moles of a compound.
8. Calculate the number of moles from a given number of molecules.
9. Calculate the number of molecules and atoms in a given mass of a compound.
10. Determine the empirical formula from percent composition.
11. Determine the molecular formula from empirical formula and molar mass.

STEP 3

Calculate the mass of magnesium reacted by subtracting the mass of the test tube from the mass of the test tube and magnesium:
Mass of Mg=22.52 g−22.42 g=0.10 g\text{Mass of Mg} = 22.52 \, \text{g} - 22.42 \, \text{g} = 0.10 \, \text{g}

STEP 4

Calculate the mass of magnesium oxide produced by subtracting the mass of the test tube from the mass of the test tube and product:
Mass of MgO=23.01 g−22.42 g=0.59 g\text{Mass of MgO} = 23.01 \, \text{g} - 22.42 \, \text{g} = 0.59 \, \text{g}

STEP 5

Calculate the mass of oxygen gas reacted by subtracting the mass of magnesium from the mass of magnesium oxide:
Mass of O2=0.59 g−0.10 g=0.49 g\text{Mass of } \mathrm{O}_{2} = 0.59 \, \text{g} - 0.10 \, \text{g} = 0.49 \, \text{g}

STEP 6

Calculate the experimental percent composition of magnesium and oxygen in magnesium oxide:
%Mg=(0.10 g0.59 g)×100=16.9%\% \text{Mg} = \left(\frac{0.10 \, \text{g}}{0.59 \, \text{g}}\right) \times 100 = 16.9\%
%O=(0.49 g0.59 g)×100=83.1%\% \text{O} = \left(\frac{0.49 \, \text{g}}{0.59 \, \text{g}}\right) \times 100 = 83.1\%

STEP 7

The theoretical percent composition of magnesium in magnesium oxide is given as 60.3%60.3\%.

STEP 8

Calculate the molar mass of zinc chloride (ZnCl2\text{ZnCl}_2) and sodium phosphate (Na3PO4\text{Na}_3\text{PO}_4):
Molar mass of ZnCl2=136.28 g/mol\text{Molar mass of ZnCl}_2 = 136.28 \, \text{g/mol}
Molar mass of Na3PO4=163.94 g/mol\text{Molar mass of Na}_3\text{PO}_4 = 163.94 \, \text{g/mol}

STEP 9

Calculate the mass of 2.36 moles of silver nitrate (AgNO3\text{AgNO}_3):
Mass=2.36 mol×170 g/mol=401.2 g\text{Mass} = 2.36 \, \text{mol} \times 170 \, \text{g/mol} = 401.2 \, \text{g}

STEP 10

Calculate the number of moles in 5.62×10275.62 \times 10^{27} molecules of sodium chloride (NaCl\text{NaCl}):
Moles=5.62×1027 molecules6.022×1023 molecules/mol=9.34×103 moles\text{Moles} = \frac{5.62 \times 10^{27} \, \text{molecules}}{6.022 \times 10^{23} \, \text{molecules/mol}} = 9.34 \times 10^{3} \, \text{moles}

STEP 11

Calculate the number of molecules and atoms in a 3.54 g sample of propane (C3H8\text{C}_3\text{H}_8):
Moles of C3H8=3.54 g44.1 g/mol=0.0803 mol\text{Moles of C}_3\text{H}_8 = \frac{3.54 \, \text{g}}{44.1 \, \text{g/mol}} = 0.0803 \, \text{mol}
Molecules=0.0803 mol×6.022×1023 molecules/mol=4.83×1022 molecules\text{Molecules} = 0.0803 \, \text{mol} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 4.83 \times 10^{22} \, \text{molecules}
Atoms=4.83×1022 molecules×11 atoms/molecule=5.31×1023 atoms\text{Atoms} = 4.83 \times 10^{22} \, \text{molecules} \times 11 \, \text{atoms/molecule} = 5.31 \times 10^{23} \, \text{atoms}

STEP 12

0 Determine the empirical formula from percent composition of calcium, phosphorus, and oxygen:
Convert the percentages to moles:
Ca: 38.8 g40.08 g/mol=0.968 mol\text{Ca: } \frac{38.8 \, \text{g}}{40.08 \, \text{g/mol}} = 0.968 \, \text{mol}
P: 20.0 g30.97 g/mol=0.646 mol\text{P: } \frac{20.0 \, \text{g}}{30.97 \, \text{g/mol}} = 0.646 \, \text{mol}
O: 41.2 g16.00 g/mol=2.575 mol\text{O: } \frac{41.2 \, \text{g}}{16.00 \, \text{g/mol}} = 2.575 \, \text{mol}
Divide by the smallest number of moles to find the ratio:
Ca: 0.9680.646≈1.5(multiply by 2 to get whole numbers)\text{Ca: } \frac{0.968}{0.646} \approx 1.5 \quad \text{(multiply by 2 to get whole numbers)}
P: 0.6460.646=1\text{P: } \frac{0.646}{0.646} = 1
O: 2.5750.646≈4\text{O: } \frac{2.575}{0.646} \approx 4
Empirical formula: Ca3P2O8\text{Ca}_3\text{P}_2\text{O}_8

STEP 13

1 Determine the molecular formula from empirical formula C6H2OCl2\text{C}_6\text{H}_2\text{OCl}_2 and molar mass 322 g/mol322 \, \text{g/mol}:
Calculate the empirical formula mass:
Empirical formula mass=6(12.01)+2(1.01)+16.00+2(35.45)=161.02 g/mol\text{Empirical formula mass} = 6(12.01) + 2(1.01) + 16.00 + 2(35.45) = 161.02 \, \text{g/mol}
Determine the multiple:
Multiple=322 g/mol161.02 g/mol=2\text{Multiple} = \frac{322 \, \text{g/mol}}{161.02 \, \text{g/mol}} = 2
Multiply the subscripts in the empirical formula by 2:
Molecular formula: C12H4O2Cl4\text{C}_{12}\text{H}_4\text{O}_2\text{Cl}_4

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