Question

A compound, \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{x}\), where \(x\) is unknown, is analyzed and found to contain \(39.70 \% \mathrm{Cr}\). What is the value of \(x\) ?

Short answer

The value of \(x\) in the compound \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{x}\) is 2, making the formula \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}\).

Step-by-step solution

Calculate the moles of each element in the compound.

We are given the information for Chromium (Cr) only, and we assume that we have 100g of the compound. Then, we have 39.7g of Cr, and the rest of the mass would be \(\mathrm{Na}_{2} \mathrm{O}_{x}\). To calculate the moles of Cr, we need to know its molar mass, which can be found in the periodic table and is approximately 52 g/mol. Therefore, the moles of Cr can be calculated as follows: Moles of Cr = \(\frac{39.7}{52} \approx 0.763\) moles

Calculate the mass of sodium and oxygen in the compound.

We know that the compound contains 39.7% of Cr, so the mass of \(\mathrm{Na}_{2} \mathrm{O}_{x}\) in the 100g sample is: \(\text{Mass of } \mathrm{Na}_{2}\mathrm{O}_{x} = 100 - 39.7 = 60.3 \text{ g}\) Since there are two moles of sodium (Na) for every one mole of Cr, we can calculate the mass of sodium in the compound using the mole ratio of \(\frac{\mathrm{Na_{2}}}{\mathrm{Cr_{2}}}\) and the moles of Cr found in step 1: Mass of Na = \(2 \times 0.763 \times \text{molar mass of Na}\) We know that the molar mass of Na is approximately 23 g/mol, so: Mass of Na = \(2 \times 0.763 \times 23 \approx 35.05 \text{ g}\) Now, we can find the mass of oxygen (O) in the compound by subtracting the total mass of Na and Cr from the mass of the whole compound: Mass of O = \(100 - 39.7 - 35.05 \approx 25.25 \text{ g}\)

Calculate the moles of oxygen in the compound.

With the mass of oxygen found, we can now calculate the moles of O in the compound using its molar mass, which is approximately 16 g/mol: Moles of O = \(\frac{25.25}{16} \approx 1.578 \text{ moles}\)

Determine the empirical formula of the compound.

Now that we have the moles of each element in the compound, we can find the empirical formula by dividing all mole values by the smallest mole value, which in this case is the moles of Cr (0.763). This will give us the mole ratio that we can round to the nearest whole number: Mole ratio of Cr = \(\frac{0.763}{0.763} = 1\) Mole ratio of Na = \(\frac{1.578}{0.763} = 2\) Mole ratio of O = \(\frac{1.578}{0.763} = 2\) The empirical formula is, therefore, \(\mathrm{Na}_{2} \mathrm{Cr}_{1} \mathrm{O}_{2}\) or \(\mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{2}\).

Determine the value of x.

Since we have found the empirical formula to be \(\mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{2}\), the value of \(x\) should be equal to 2. Hence, the correct formula for the compound is \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}\).

Key concepts

Molar Mass

The molar mass of an element is the mass of one mole of its atoms. It's typically expressed in grams per mole (g/mol). We often use the molar mass in chemical calculations to convert between the mass of a substance and the amount in moles. This relationship is crucial for analyzing compounds and understanding their composition.
The periodic table provides the molar mass for each element. For instance, the molar mass of Chromium (Cr) is around 52 g/mol, and Sodium (Na) is about 23 g/mol. These values are averages of the isotopic masses of an element because naturally occurring elements are mixtures of isotopes.

• Understanding molar mass is essential for calculating the amount of each element in a compound.
• It helps convert grams to moles, aiding in empirical formula derivation.

In our exercise, we used the molar mass of Cr to determine its role in the compound and ensure accurate stoichiometric calculations.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. This branch of chemistry uses mole concepts and ratios to calculate amounts within a chemical equation.
It involves the use of coefficients from balanced equations and molar masses to deduce unknown quantities in a reaction.

• Stoichiometry helps determine the proportions of elements within compounds.
• It uses the mole concept to balance and analyze chemical strategies efficiently.

In the given exercise, stoichiometry aided in figuring out the distribution of sodium, chromium, and oxygen in the compound \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{x}\). For instance, understanding that two moles of sodium correspond to one mole of chromium helps streamline our calculations and ensure accuracy in deriving the empirical formula of the compound.

Chemical Composition Analysis

Chemical composition analysis is the process of identifying the elements within a compound and quantifying them. It allows us to determine the percentage of each element present, which is vital for understanding the compound's properties and chemical behavior.
By analyzing the chemical composition, we can deduce ratios, empirical formulas, and evaluate purity. This process is fundamental in laboratories to ensure substances meet required specifications for chemical reactions and manufacturing.

• Empirical formula calculation is one outcome of chemical composition analysis.
• It involves determining the relative proportions of different elements in a compound.

In our problem, chemical composition analysis helped us determine the amount of chromium present and consequently how many oxygen atoms (\(x\)) are part of the compound \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{x}\). The exercises guided us through understanding which elements were present and in which ratios, aligning with the given percentage composition of chromium to solve for \(x\).