Stoichiometry: Mass, Moles & Concentration

To find the number of moles of calcium carbonate (CaCO₃), divide the mass of the sample by its molar mass. The formula is: moles = mass (g) / molar mass (g/mol). Here, the mass is 50.0 g, and the molar mass of CaCO₃ is 100.09 g/mol. Performing the calculation: 50.0 g / 100.09 g/mol = 0.4995 mol, which rounds to 0.500 mol. This indicates that there are 0.500 moles of CaCO₃ in the sample.

Percent yield is calculated by dividing the actual yield by the theoretical yield and then multiplying by 100. In this case, the actual yield is 68.0 g and the theoretical yield is 85.0 g. The calculation is (68.0 g / 85.0 g) × 100, which equals 80.0%. This percentage indicates the efficiency of the reaction, showing that 80% of the expected product was successfully produced.

To find the mass percent of nitrogen in ammonium nitrate (NH₄NO₃), first calculate the molar mass of the compound. NH₄NO₃ consists of two nitrogen atoms (2 × 14.01 g/mol), four hydrogen atoms (4 × 1.008 g/mol), and three oxygen atoms (3 × 16.00 g/mol). The total molar mass of NH₄NO₃ is 80.04 g/mol. The mass of nitrogen in the compound is 28.02 g/mol (from the two nitrogen atoms). The mass percent of nitrogen is then (28.02 g/mol / 80.04 g/mol) × 100%, which equals 35.00%.

To find the volume of the stock solution needed, we can use the dilution equation M₁V₁ = M₂V₂. Here, M₁ is the molarity of the stock solution (12.0 M), M₂ is the desired molarity (0.500 M), and V₂ is the final volume (500.0 mL). Rearranging the equation to solve for V₁ (the volume of the stock solution), we get V₁ = (M₂V₂) / M₁. Substituting the values gives V₁ = (0.500 M × 500.0 mL) / 12.0 M, which calculates to 20.8 mL.

To find the number of molecules in 3.50 moles of water, multiply the number of moles by Avogadro's number (6.02 × 10²³). This calculation is done as follows: 3.50 moles × 6.02 × 10²³ molecules/mole = 2.11 × 10²⁴ molecules. This result indicates that there are approximately 2.11 sextillion molecules of water in 3.50 moles, demonstrating the relationship between moles and the number of molecules in chemistry.

Percent error is calculated using the formula: \(\text{Percent Error} = \left(\frac{\text{Absolute Error}}{\text{Actual Value}}\right) \times 100\). Here, the absolute error is the difference between the measured value (9.80 g) and the actual value (10.00 g), which is 0.20 g. Plugging the values into the formula gives: \(\left(\frac{0.20}{10.00}\right) \times 100 = 2.00\%\). This indicates that the measured value deviates by 2.00% from the actual value, representing the accuracy of the measurement.

To calculate the molarity of a solution, use the formula: Molarity (M) = moles of solute / liters of solution. Here, 0.750 moles of NaCl are dissolved in 250.0 mL of solution, which is 0.250 liters. Therefore, the molarity is calculated as 0.750 moles / 0.250 L = 3.00 M. This indicates that there are 3.00 moles of NaCl in every liter of the solution, reflecting a relatively concentrated solution.

To calculate the mass of glucose required for a 0.250 M solution in 2.00 L, use the formula: mass (g) = molarity (mol/L) × volume (L) × molar mass (g/mol). First, find the number of moles: 0.250 mol/L × 2.00 L = 0.500 moles. Then, multiply by the molar mass of glucose: 0.500 moles × 180.16 g/mol = 90.08 g. Rounding this to one decimal place gives 90.1 g, which is the mass needed to prepare the solution.