Question

Calculate the mass in grams of each of the following samples. a. \(2.6 \times 10^{-2}\) mol of copper(II) sulfate, \(\mathrm{CuSO}_{4}\) b. \(3.05 \times 10^{3}\) mol of tetrafluoroethylene, \(\mathrm{C}_{2} \mathrm{F}_{4}\) c. 7.83 mmol \((1 \mathrm{mmol}=0.001 \mathrm{mol})\) of 1,4 pentadiene, \(\mathrm{C}_{5} \mathrm{H}_{8}\). d. 6.30 mol of bismuth trichloride, \(\mathrm{BiCl}_{3}\) e. 12.2 mol of sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)

Short answer

a. 4.15 g of CuSO4 b. 3.05 × 10^5 g of C2F4 c. 0.53 g of C5H8 d. 1986.08 g of BiCl3 e. 4176.06 g of C12H22O11

Step-by-step solution

a. Copper(II) sulfate, CuSO4

1. Determine the molar mass of CuSO4: Molar mass of CuSO4 = mass of Cu + mass of S + (4 * mass of O) Using the atomic masses from the periodic table, Molar mass of CuSO4 = 63.5 g/mol + 32.07 g/mol + (4 * 16 g/mol) Molar mass of CuSO4 = 159.57 g/mol 2. Calculate the mass of \(2.6 \times 10^{-2}\) mol of CuSO4: Mass = moles * molar mass Mass = \(2.6 \times 10^{-2}\) mol * 159.57 g/mol Mass = 4.15 g

b. Tetrafluoroethylene, C2F4

1. Determine the molar mass of C2F4: Molar mass of C2F4 = (2 * mass of C) + (4 * mass of F) Using the atomic masses from the periodic table, Molar mass of C2F4 = (2 * 12.01 g/mol) + (4 * 19 g/mol) Molar mass of C2F4 = 100.02 g/mol 2. Calculate the mass of \(3.05 \times 10^{3}\) mol of C2F4: Mass = moles * molar mass Mass = \(3.05 \times 10^{3}\) mol * 100.02 g/mol Mass = 3.05 × 10^5 g

c. 1,4 pentadiene, C5H8

1. Determine the molar mass of C5H8: Molar mass of C5H8 = (5 * mass of C) + (8 * mass of H) Using the atomic masses from the periodic table, Molar mass of C5H8 = (5 * 12.01 g/mol) + (8 * 1.01 g/mol) Molar mass of C5H8 = 68.09 g/mol 2. Calculate the mass of 7.83 mmol of C5H8 (1 mmol = 0.001 mol): Moles = 7.83 mmol * 0.001 mol/mmol = 0.00783 mol Mass = moles * molar mass Mass = 0.00783 mol * 68.09 g/mol Mass = 0.53 g

d. Bismuth trichloride, BiCl3

1. Determine the molar mass of BiCl3: Molar mass of BiCl3 = mass of Bi + (3 * mass of Cl) Using the atomic masses from the periodic table, Molar mass of BiCl3 = 208.98 g/mol + (3 * 35.45 g/mol) Molar mass of BiCl3 = 315.33 g/mol 2. Calculate the mass of 6.30 mol of BiCl3: Mass = moles * molar mass Mass = 6.30 mol * 315.33 g/mol Mass = 1986.08 g

e. Sucrose, C12H22O11

1. Determine the molar mass of C12H22O11: Molar mass of C12H22O11 = (12 * mass of C) + (22 * mass of H) + (11 * mass of O) Using the atomic masses from the periodic table, Molar mass of C12H22O11 = (12 * 12.01 g/mol) + (22 * 1.01 g/mol) + (11 * 16 g/mol) Molar mass of C12H22O11 = 342.30 g/mol 2. Calculate the mass of 12.2 mol of C12H22O11: Mass = moles * molar mass Mass = 12.2 mol * 342.30 g/mol Mass = 4176.06 g

Key concepts

Molar Mass

Molar mass serves as a bridge between the mass of a substance and the amount of substance in moles. It is defined as the mass of one mole of a given substance and is expressed in grams per mole (g/mol). To find the molar mass, you add together the atomic masses of all the atoms in a molecule. For instance, to calculate the molar mass of water (H₂O), you would sum the masses of two hydrogen atoms and one oxygen atom.

For accuracy in mole to mass calculations, you look at a compound's chemical formula to ascertain how many atoms of each element are present. You then multiply the atomic mass of each element by its count within the molecule and sum these values. This gives us the total molar mass for the compound.

Stoichiometry

Stoichiometry is the field of chemistry that relates to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the law of conservation of mass where the mass of the reactants equals the mass of the products in a chemical reaction. It uses molar mass as a conversion factor to relate masses of substances to the amounts in moles. This is critical when deciphering reaction yields or when you need to scale reactions up or down.

In practice, stoichiometry allows for the calculation of how much product can be made from a given amount of reactants (theoretical yield) or how much reactant is required to produce a certain amount of product. Understanding stoichiometry is fundamental in developing a strong foundation in chemistry.

Chemical Formulas

Chemical formulas give us a concise way to represent the elements and their quantities within a chemical compound. For example, the formula for glucose is C₆H₁₂O₆, indicating that each molecule contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.

When performing mole to mass calculations, chemical formulas play a crucial role as they guide you in determining the correct amount of each element to consider for the molar mass. A balance of equations also relies on these formulas, as they ensure that atoms are conserved in the reaction, which is consistent with the principle of stoichiometry.

Atomic Masses

Atomic masses are the weights of individual atoms, usually measured in atomic mass units (amu). In a broader and more practical sense for chemistry calculations, you use the average atomic mass which is the weighted average of the isotopes of an element as they naturally occur. This value is what's published on the periodic table and is critical when calculating molar masses for substances.

To illustrate, the atomic mass of carbon is roughly 12.01 g/mol, which accounts for the natural abundance of its isotopes, predominantly Carbon-12 and Carbon-13. This value is what you would use when calculating the molar mass of compounds containing carbon, like in the computation of the molar mass of glucose mentioned earlier.