Question

How many moles of compound are in \(25.0 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) ?

Short answer

0.0874 moles

Step-by-step solution

Find the Molar Mass of the Compound

To find the molar mass of \(\text{Na}_{2} \text{CO}_{3} \cdot 10 \text{H}_{2} \text{O}\), add the molar masses of all the atoms in the formula. This includes 2 sodium (Na) atoms, 1 carbon (C) atom, 3 oxygen (O) atoms for \(\text{Na}_{2} \text{CO}_{3}\), and 10 water (H\textsubscript{2}O) molecules. Use the periodic table for the atomic masses.

Calculate the Molar Mass

The atomic masses are: \(\text{Na}\) = 22.99 g/mol, \( \text{C} \) = 12.01 g/mol, \(\text{O}\) = 16.00 g/mol, and \(\text{H}\) = 1.01 g/mol. Calculate: \text{Na}_{2} \text{CO}_{3} = (2 x 22.99) + 12.01 + (3 x 16.00) = 105.99 g/mol Water: (10 x (2 x 1.01 + 16.00)) = 180.20 g/mol Total Molar Mass: 105.99 g/mol + 180.20 g/mol = 286.19 g/mol.

Convert Grams to Moles

Use the relationship: \[ \text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \]. For \(25.0 \text{ g of }\text{Na}_{2} \text{CO}_{3} \cdot 10 \text{H}_{2} \text{O}\): \[ \text{moles} = \frac{25.0 \text{ g}}{286.19 \text{ g/mol}} = 0.0874 \text{ moles} \]

Key concepts

molar mass

Understanding molar mass is essential in chemistry; it is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. To find the molar mass, you add up the atomic masses of all the atoms in the compound's formula. For example, let's find the molar mass of \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\) (Sodium Carbonate Decahydrate). You begin with the atomic masses: \(\text{Na}\) is 22.99 g/mol, \(\text{C}\) is 12.01 g/mol, \(\text{O}\) is 16.00 g/mol, and \(\text{H}\) is 1.01 g/mol. So, the calculation is:
\(\text{Na}_2 \text{CO}_3 : (2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99 \text{ g/mol}\).
Water: (10 \times (2 \times 1.01 + 16.00)) = 180.20 \text{ g/mol}.
Total molar mass of \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\) = 105.99 g/mol + 180.20 g/mol = 286.19 g/mol.

stoichiometry

Stoichiometry is all about the quantitative relationships in chemical reactions. It involves using ratios derived from balanced chemical equations to calculate the amounts of reactants and products. In our exercise, stoichiometry helps us understand how many moles of sodium carbonate decahydrate (\(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\)) we have when given a specific mass. The step-by-step calculation shows how stoichiometry enables the conversion of mass to moles using the molar mass as a bridge, making it a crucial tool in solving chemical equations.

moles to grams conversion

Converting between moles and grams is a fundamental skill in chemistry. In the exercise, we use the relationship: \(\text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \), to convert the given mass of \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\). For instance, for 25.0 grams of the compound, the calculation is:
\(\text{moles} = \frac{25.0 \text{ grams}}{286.19 \text{ g/mol}} = 0.0874 \text{ moles} \). This conversion is straightforward: divide the mass by the molar mass to find the number of moles. Understanding this concept is vital for accurate chemical calculations and ensures precise measurements in laboratory settings.

chemical formulas

Chemical formulas represent the types and numbers of atoms in a molecule of a compound. They provide essential information needed to compute molar masses, perform stoichiometric calculations, and understand the structure of the compound. For instance, the formula \(\text{Na}_2 \text{CO}_3 \cdot 10 \text{H}_2 \text{O}\) reveals that each molecule contains 2 sodium (Na) atoms, 1 carbon (C) atom, and 3 oxygen (O) atoms, plus 10 water (H\(_2\)O) molecules. Such formulas are indispensable in chemical analysis and synthesis, as they encapsulate the proportions and relationships between different atoms in a compound, helping chemists to predict reaction outcomes and to design new substances.