Question

Calculate the mass in grams of the following. (a) \(1.25 \mathrm{~mol} \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (b) \(0.600 \mu \mathrm{mol} \mathrm{C}_{4} \mathrm{H}_{10}\) (c) \(0.625 \mathrm{mmol}\) iron(III) nitrate (d) 1.45 mol ammonium carbonate

Short answer

The mass in grams of (a) 1.25 mol Ca3(PO4)2 is 310.18 g, (b) 0.600 \(\mu\)mol C4H10 is 0.000072 g, (c) 0.625 mmol iron(III) nitrate is 0.104 g, (d) 1.45 mol ammonium carbonate is 144.42 g.

Step-by-step solution

Calculate Molar Mass of Ca3(PO4)2

Find molar mass by adding the mass of 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms. The molar mass of calcium (Ca) is 40.08 g/mol, phosphorus (P) is 30.97 g/mol, and oxygen (O) is 16.00 g/mol. Thus, the molar mass of Ca3(PO4)2 is calculated as: 3(40.08 g/mol) + 2(30.97 g/mol) + 8(16.00 g/mol).

Convert Moles to Mass for Ca3(PO4)2

Multiply the number of moles of Ca3(PO4)2 by its molar mass to convert moles to grams. Use the molar mass from the previous step and the given amount of substance in moles.

Calculate Molar Mass of C4H10

Find molar mass by adding the mass of 4 carbon atoms and 10 hydrogen atoms. The molar mass of carbon (C) is 12.01 g/mol, and hydrogen (H) is 1.008 g/mol. Thus, the molar mass of C4H10 is calculated as: 4(12.01 g/mol) + 10(1.008 g/mol).

Convert Micro-moles to Grams for C4H10

Multiply the number of micro-moles of C4H10 by its molar mass to convert to micro-grams, and then convert micro-grams to grams (1 gram = 1,000,000 micrograms).

Calculate Molar Mass of Iron(III) Nitrate

Iron(III) nitrate has a formula of Fe(NO3)3. The molar mass is found by adding the mass of 1 iron atom, 3 nitrogen atoms, and 9 oxygen atoms. The molar mass of iron (Fe) is 55.85 g/mol, nitrogen (N) is 14.01 g/mol, and oxygen (O) is 16.00 g/mol. Thus, calculate the molar mass as: 1(55.85 g/mol) + 3(14.01 g/mol) + 9(16.00 g/mol).

Convert Milli-moles to Grams for Iron(III) Nitrate

Multiply the number of milli-moles of iron(III) nitrate by its molar mass to convert to milli-grams, and then convert milli-grams to grams (1 gram = 1,000 milligrams).

Calculate Molar Mass of Ammonium Carbonate

Ammonium carbonate has a chemical formula of (NH4)2CO3. The molar mass is found by adding the mass of 2 nitrogen atoms, 8 hydrogen atoms, 1 carbon atom, and 3 oxygen atoms. Use the molar mass of nitrogen (N) 14.01 g/mol, hydrogen (H) 1.008 g/mol, carbon (C) 12.01 g/mol, and oxygen (O) 16.00 g/mol to find the molar mass.

Convert Moles to Mass for Ammonium Carbonate

Multiply the number of moles of ammonium carbonate by its molar mass to convert moles to grams. Use the molar mass from the previous step and the given amount of substance in moles.

Key concepts

Mole to Gram Conversion

Understanding how to convert moles to grams is foundational in chemistry for quantifying substances. A mole is a unit that represents a fixed number of particles, typically atoms or molecules, and is set at Avogadro's number, which is approximately
\(6.022 \times 10^{23}\) particles. The connection between moles and grams is bridged by the substance's molar mass, which is the mass of one mole of its particles expressed in grams.
The conversion is straightforward: take the number of moles of the substance and multiply it by the substance's molar mass. This process was applied in the provided solutions to calculate the mass of different chemicals. It's important to remember units of measure during conversion, especially when dealing with micro (\br)\(10^{-6}\)) or milli (\br)\(10^{-3}\)) moles, as shown in step 4 and step 6 of the problem solution.

Chemical Formula

A chemical formula acts as a shorthand representation of the composition and proportion of atoms within a compound. Elements in the formula are denoted by their chemical symbols, such as
Ca for calcium, while subscripts indicate the number of each type of atom in the molecule. For example, \(Ca_3(PO4)_2\) represents a compound with 3 calcium atoms, 2 phosphorus atoms, and 8 oxygen atoms. Understanding the components of a chemical formula is essential for tasks such as calculating molar mass, as demonstrated in the solutions provided for steps 1, 3, 5, and 7. A clear comprehension of chemical formulas also aids in the understanding of reaction stoichiometry, as it provides the ratio of reactants and products involved in a chemical reaction.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of the mole. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a reaction, making it crucial for everything from lab experiments to industrial processes.
In stoichiometry, the coefficients in a balanced chemical equation indicate the ratio of moles of each substance involved. For instance, if a reaction states that 1 mole of hydrogen reacts with 1 mole of oxygen to produce 2 moles of water (\(2H_2 + O_2 \rightarrow 2H_2O\)), stoichiometry tells us that 2 moles of hydrogen gas will react completely with 1 mole of oxygen gas. This balanced equation is central to solving problems related to reaction yields and reactant conversion, though it was not the primary focus of the exercise provided.

Atomic and Molecular Mass

Atomic and molecular mass are fundamental concepts in chemistry that relate to the mass of individual atoms and molecules. The atomic mass of an element, often found on the periodic table, is approximately equal to the number of protons and neutrons in the nucleus and is expressed in atomic mass units (amu). One atomic mass unit is defined as one-twelfth the mass of a carbon-12 atom.
Molecular mass, also referred to as molecular weight, is the sum of the atomic masses of all atoms in a molecule. For complex substances, you calculate it by considering the quantity and types of atoms present as was done in the textbook solutions (steps 1, 3, 5, and 7). The molecular mass gives the molar mass of the substance in grams per mole (g/mol), a critical value used in mole to gram conversion. Accurate determination of atomic and molecular mass forms the basis for many calculations in the field of chemistry, including those performed in the initial problem.