Question

A mixture of \(\mathrm{NaBr}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) contains 29.96 percent Na by mass. Calculate the percent by mass of each compound in the mixture.

Short answer

NaBr: 24.1% and Na2SO4: 75.9% by mass.

Step-by-step solution

Define Known Quantities and Variables

Let's start by defining what we know. - The mixture consists of NaBr and \( \mathrm{Na}_2\mathrm{SO}_4 \).- Let \( x \) be the mass fraction of NaBr in the mixture, and \( (1-x) \) be the mass fraction of \( \mathrm{Na}_2\mathrm{SO}_4 \). - We need to find x and \( 1-x \).

Write the Formula for Sodium Content

The sodium content can be expressed in terms of the mass fractions:\[\text{Percent Na} = x \times \frac{\text{Mass of Na in NaBr}}{\text{Molar Mass of NaBr}} + (1-x) \times \frac{\text{Mass of Na in } \mathrm{Na}_2\mathrm{SO}_4}{\text{Molar Mass of } \mathrm{Na}_2\mathrm{SO}_4}\]

Calculate Molar Masses and Na Content

- **NaBr**: Molar mass is 102.89 g/mol with 22.99 g/mol contributed by Na.- **Na\(_2\)SO\(_4\)**: Molar mass is 142.04 g/mol with 45.98 g/mol contributed by Na (as it has 2 Na atoms). Using these, the formula becomes:\[29.96 = x \times \frac{22.99}{102.89} \times 100 + (1-x) \times \frac{45.98}{142.04} \times 100\]

Substitute Values and Solve for x

Substitute the known values into the formula:\[29.96 = x \times 22.35 + (1-x) \times 32.38\]Solve for \( x \):- Rearrange to get: \( 29.96 = 22.35x + 32.38 - 32.38x \)- Simplify: \( 29.96 = 32.38 - 10.03x \)- Solve for \( x \): \( x = \frac{32.38 - 29.96}{10.03} = 0.241 \)

Calculate Percent by Mass for Each Compound

Now compute the percent by mass of each compound:- **NaBr** (x): \( x \times 100 = 0.241 \times 100 = 24.1 \% \)- **Na\(_2\)SO\(_4\)** (1-x): \( (1-x) \times 100 = (1-0.241) \times 100 = 75.9\% \)

Key concepts

Molar Mass

Molar mass is a crucial concept in chemistry that helps us understand the weight of one mole of a given substance. It essentially acts as a bridge between the microscopic world of atoms and the macroscopic world of grams and kilograms. When we talk about molar mass, we are referring to the mass of one mole of an element or compound expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you simply add up the atomic masses of all the atoms in a chemical formula. Atomic masses can be found on the periodic table and are usually expressed in unified atomic mass units (u), but when calculating molar mass, the same number is expressed in grams.For example:- The molar mass of sodium bromide (NaBr) is calculated by adding the atomic masses of sodium (Na) and bromine (Br). Given: Na = 22.99 u and Br = 79.90 u, the molar mass of NaBr is 102.89 g/mol.- Similarly, for sodium sulfate ( \(\mathrm{Na}_{2}\mathrm{SO}_{4} \)), which contains two sodium, one sulfur, and four oxygen atoms, you'd calculate based on Na = 22.99 u, S = 32.07 u, and O = 16.00 u.Understanding molar mass is essential because it allows chemists to convert between the number of atoms or molecules and the mass of a sample, which is indispensable when dealing with chemical reactions and compositions.

Sodium Content

When analyzing a compound or mixture, it's often important to determine how much of a particular element is present, such as sodium (Na). The sodium content refers to the proportion of sodium by mass in a compound or mixture compared to its total mass. This information is especially critical in mixtures consisting of multiple compounds where sodium is a key component, as seen in food products or chemical reactions.To calculate sodium content in compounds like NaBr and Na\(_2\)SO\(_4\), you follow these steps:- Determine how many sodium atoms are in the chemical formula. For example, NaBr has one sodium atom, while Na\(_2\)SO\(_4\) has two.- Calculate the molar mass of the compound.- Find the total mass of the sodium atoms in the compound.For examplanation,- In NaBr, the sodium contributes 22.99 g/mol out of its total 102.89 g/mol.- For Na\(_2\)SO\(_4\), sodium contributes 45.98 g/mol (as it contains two sodium atoms) from its total 142.04 g/mol.This information is then expressed as a percentage to indicate the sodium content by mass, which is essential for calculating the mass percent of sodium in mixtures.

NaBr

Sodium bromide (\(\mathrm{NaBr}\)) is an ionic compound comprised of sodium ions (Na\(^+\)) and bromide ions (Br\(^-\)). It is a white, crystalline solid at room temperature, and is highly soluble in water.The properties of NaBr make it useful in various applications:- **Pharmaceuticals:** NaBr has been used historically as a sedative and anticonvulsant. - **Chemical Synthesis:** In the laboratory, it is used as a source of bromide ions.- **Industrial Processes:** NaBr is used in oil drilling fluids.When working with NaBr in chemical calculations, it's important to remember its molar mass is 102.89 g/mol. To understand its sodium content focusing on its single sodium atom:- The sodium ion is responsible for 22.99 g/mol of the compound's mass.This calculation is helpful especially when determining the sodium content in a mixture, as it directly correlates with the molar mass and application of the substance.

Na2SO4

Sodium sulfate (\(\mathrm{Na}_{2}\mathrm{SO}_{4}\)) is another important ionic compound, composed of two sodium ions (Na\(^+\)) and a sulfate ion (SO\(_4\)\(^2-\)). It typically appears as a white crystalline solid and is known for its good solubility in water.Na\(_2\)SO\(_4 \) is often used in:- **Detergents and Soaps:** It acts as a filler to improve cleaning agents' consistency.- **Paper and Glass Manufacturing:** Serves as a drying agent in these industries.- **Laboratories:** Used as a standard for volumetric analysis.The molar mass of Na\(_2\)SO\(_4\) is calculated as 142.04 g/mol. This compound's composition includes:- Two sodium atoms contributing 45.98 g/mol to the molar mass.Understanding the composition of sodium in Na\(_2\)SO\(_4\) is essential for determining its sodium percentage in mixtures. This aids in understanding its role either as a reactant or an ingredient in various mixtures or formulations.