Question

5 multiple choice 10 points determine the mass percent of c in sodium bicarbonate (nahco₃) 14.28% 50.00% 28.56% 7.14% 6 multiple choice 10 points a compound was found to contain 90.6% lead (pb) and 9.4% oxygen. the empirical formula for this compound is pb₂o₄ pbo pb₂o₃ pbo₂

Explanation:

5.

Step1: Calculate molar mass of NaHCO₃

The molar mass of Na (sodium) is approximately 22.99 g/mol, H (hydrogen) is approximately 1.01 g/mol, C (carbon) is approximately 12.01 g/mol, and O (oxygen) is approximately 16.00 g/mol. For NaHCO₃, \(M = 22.99+1.01 + 12.01+3\times16.00=84.01\) g/mol.

Step2: Calculate mass - percent of C

The mass - percent of C is \(\frac{12.01}{84.01}\times100\%\approx14.28\%\)

Step1: Assume 100 g of the compound

If we assume 100 g of the compound, then the mass of lead (Pb) is 90.6 g and the mass of oxygen (O) is 9.4 g.

Step2: Calculate moles of each element

The molar mass of Pb is approximately 207.2 g/mol and of O is 16.00 g/mol. Moles of Pb, \(n_{Pb}=\frac{90.6}{207.2}\approx0.437\) mol. Moles of O, \(n_O=\frac{9.4}{16.00}\approx0.5875\) mol.

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles (\(n_{Pb} = 0.437\) mol). For Pb, \(\frac{0.437}{0.437}=1\). For O, \(\frac{0.5875}{0.437}\approx1.35\). Multiply by 3 to get whole - number ratios. We get \(Pb_3O_4\).

Answer:

14.28%

6.