Question

Sodium borate decahydrate, \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) is commonly known as borax. It is used as a deodorizer and mold inhibitor. A sample weighing \(15.86 \mathrm{~g}\) is heated until a constant mass is obtained indicating that all the water has been evaporated off. (a) What percent, by mass of \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) is water? (b) What is the mass of the anhydrous sodium borate, \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} ?\)

Short answer

Answer: The percentage of water by mass in sodium borate decahydrate is approximately 62.87%. The mass of anhydrous sodium borate in the given sample is approximately 5.88 g.

Step-by-step solution

1. Determine the molar mass of Sodium borate decahydrate

To find the molar mass, we need to add the molar masses of all the atoms in the compound. Molar mass of \(\mathrm{Na} = 23\) g/mol, Molar mass of \(\mathrm{B} = 10.81\) g/mol, Molar mass of \(\mathrm{O} = 16\) g/mol, and Molar mass of \(\mathrm{H}_{2}\mathrm{O} = 18\) g/mol. The molar mass of \(\mathrm{Na}_{2} \mathrm{~B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) will be: = \(2(23) + 4(10.81) + 7(16) + 10(18)\) = \(106.24\)

2. Calculate the mass of water in the sample

Converting water molecules into mass, we get: \(\mathrm{Mass~of~water~in~Na_2B_4O_7}\cdot 10\mathrm{H_2O}= 10 \times 18\) \(\implies 180~g/mol\)

3. Determine the percentage of water by mass in Sodium borate decahydrate

Now, we can calculate the percentage of water in the sample: \(\mathrm{Percentage~of~water} = \frac{\mathrm{mass~of~water}}{\mathrm{molar~mass~of~Na_2B_4O_7}\cdot 10 \mathrm{H_2O}} * 100\%\) \(\implies \frac{180}{106.24+180} * 100\%\) \(\implies 62.87\%\) So, the percentage of water in the sample is approximately \(62.87\%\).

4. Calculate the mass of anhydrous sodium borate

Given the mass of the sample (\(15.86~g\)), we can calculate the mass of anhydrous sodium borate as: \(\mathrm{Mass~of~anhydrous~sodium~borate} = (1 - \mathrm{percentage~of~water}) \times \mathrm{mass~of~sample}\) \(\implies (1 - 0.6287) \times 15.86\) \(\implies 5.88~g\) Hence, the mass of anhydrous sodium borate in the given sample is approximately \(5.88~g\).