Problem 73 - Chemistry: An Atoms-Focused Approach

Question

What is the mass of 0.122 mol of \(\mathrm{Mg} \mathrm{CO}_{3} ?\)

Step-by-Step Solution

Answer

Answer: The mass of 0.122 mol of MgCO3 is 10.29 grams.

Step 1: Identify the molar mass of \(\mathrm{Mg} \mathrm{CO}_{3}\)

To find the molar mass of \(\mathrm{Mg} \mathrm{CO}_{3}\), we will sum the molar masses of each individual element present. Molar mass of \(\mathrm{Mg} = 24.30 \, g/mol\) Molar mass of \(\mathrm{C} = 12.01 \, g/mol\) Molar mass of \(\mathrm{O} = 16.00 \, g/mol\) Now, to determine the molar mass of \(\mathrm{MgCO}_{3}\), we sum the molar masses, remembering that there are three oxygen atoms. Molar mass of \(\mathrm{MgCO}_{3} = 24.30 + 12.01 + (3 × 16.00) = 24.30 + 12.01 + 48.00 = 84.31 \, g/mol\)

Step 2: Calculate the mass of 0.122 mol of \(\mathrm{Mg} \mathrm{CO}_{3}\)

Now that we have the molar mass of \(\mathrm{Mg} \mathrm{CO}_{3}\), we will multiply the given number of moles (0.122 mol) by the molar mass to find the mass of 0.122 mol of \(\mathrm{Mg} \mathrm{CO}_{3}\). Mass = moles × molar mass Mass = 0.122 \, mol × 84.31 \, g/mol = 10.29 \, g The mass of 0.122 mol of \(\mathrm{Mg} \mathrm{CO}_{3}\) is 10.29 grams.

Key Concepts

Moles to Grams Conversion

Converting from moles to grams is a core concept in chemistry, making it easier to understand the quantity of a substance in tangible terms. This conversion relies heavily on the molar mass of a compound, which serves as the bridge between the abstract world of molecules and the more concrete world of grams we can weigh.

• To begin with, you would need the molar mass of the compound. For example, in the case of magnesium carbonate (\( \mathrm{MgCO}_3 \)), you have calculated the molar mass to be 84.31 g/mol.
• Next, use the formula: \[\text{Mass} = \text{moles} \times \text{molar mass}\]
• In simpler terms, just multiply the number of moles you have by the molar mass of your compound.

Following this formula, if you have 0.122 moles of \( \mathrm{MgCO}_3 \), you can find the mass by multiplying it by the molar mass:\[0.122 \, \text{mol} \times 84.31 \, \text{g/mol} = 10.29 \, \text{g}\]This simple multiplication shows you how much the compound would weigh on a scale.

Chemical Compounds

Chemical compounds are substances formed from two or more different elements that are bonded together. These combinations result in a unique set of properties for each compound. Understanding chemical compounds allows chemists to predict interactions and recognize substances.

• The atoms in a chemical compound are held together by chemical bonds, sharing or transferring electrons to achieve stability.
• Each compound can be represented by a chemical formula which indicates the types and numbers of atoms involved. For instance, \( \mathrm{MgCO}_3 \) denotes that each molecule includes one magnesium atom, one carbon atom, and three oxygen atoms.

Knowing the chemical formula is key for any calculations involving molar mass or reacting ratios, as each element contributes its own atomic mass to the total molar mass of the compound.

Molar Mass of Magnesium Carbonate

The molar mass of a compound is the mass in grams of one mole of its molecules or formula units. It is the sum of the weights of all the atoms that make up a single molecule of the compound. For magnesium carbonate (\( \mathrm{MgCO}_3 \)), finding the molar mass helps in various calculations, such as converting moles to grams.
First, identify the atomic masses of the individual elements from the periodic table:

• Magnesium (\( \mathrm{Mg} \)) has an atomic mass of 24.30 g/mol.
• Carbon (\( \mathrm{C} \)) has an atomic mass of 12.01 g/mol.
• Oxygen (\( \mathrm{O} \)) has an atomic mass of 16.00 g/mol.

Next, calculate the total molar mass by adding them together, keeping in mind the number of each type of atom:\[\text{Molar mass of } \mathrm{MgCO}_3 = 24.30 + 12.01 + (3 \times 16.00) = 84.31 \, \text{g/mol}\]This calculation sums up the contribution of each element, reflecting the compound's total mass per mole, aiding in every molecular-scale calculation you undertake.