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Mole to mass:<br />
1. Determine how many moles are given in the problem<br />
2. Calculate the molar mass of the substance<br />
3. Multiply the mass by the moles given<br />
Percent Yield:<br />
1. Balance the chemical equation<br />
2. Find the limiting reagent<br />
3. FInd the theoretical yield<br />
4. Find the actual yield<br />
5. Round using significant figures<br />
6, Find the percentage yield<br />
Mole to mole problem -<br />
2 H2 + O2 --> 2 H2O<br />
6 mol H2 1 mol O2<br />
2 mol H2<br />
= 3 moles of O2<br />
Mole to gram problem -<br />
2 C + 2 H2O --> CH4 + CO2<br />
2 mol C 12 g C<br />
1 mol C = 24 g C<br />
Limiting Reagant problem -<br />
CuCI2 + 2 NaNO3 ---> Cu(NO3)2 + 2 NaCI<br />
15 g CuCI2 1 mol CuCI2 2 mol NaCI 58.5 g NaCI<br />
Percent yield formula -<br />
133.5 g CuCI2 1 mol CuCI2 1 mol NaCI<br />
21 g NaNO3 1 mol NaNO3 2 mol NaCI 58.5 g NaCI<br />
85 g NaNO3 2 mol NaNO3 1 mol NaCI<br />
Mole to mass problem -<br />
3 mol CO2 44 g CO2<br />
1 mol CO2<br />
= 132 g CO2<br />
= 13.146 g NaCI<br />
= 14.45 g NaCI<br />
Mass of actual yield / mass of theoretical yield x 100%<br />
107
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