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 \(x\) ?

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

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 = \(100 - 39.7 - 35.05 \approx 25.25 \text{ g}\)

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 = \(\frac{25.25}{16} \approx 1.578 \text{ moles}\)

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 \(\mathrm{Na}_{2} \mathrm{Cr} \mathrm{O}_{2}\).

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 \(\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}\).