Problem 187 - Chemistry Matter and Change

Question

Borax Hydrated sodium tetraborate \(\left(\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7} \times x \mathrm{H}_{2} \mathrm{O}\right)\) is commonly called borax. Chemical analysis indicates that this hydrate is 52.8\(\%\) sodium tetraborate and 47.2\(\%\) water. Determine the formula and name the hydrate.

Step-by-Step Solution

Answer

The formula is \(\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}\); it's called sodium tetraborate decahydrate.

Step 1: Understand the Components

First, note the composition of the compound: it's 52.8% sodium tetraborate (\(\text{Na}_2\text{B}_4\text{O}_7\)) and 47.2% water (\(\text{H}_2\text{O}\)).)

Step 2: Assume a Sample Mass

Assume a 100 g sample of borax. This simplifies the calculation, as you have 52.8 g of sodium tetraborate and 47.2 g of water in this 100 g sample.

Step 3: Convert Sodium Tetraborate Mass to Moles

Calculate the number of moles of sodium tetraborate. Use its molar mass, 201.22 g/mol:\[\text{moles of Na}_2\text{B}_4\text{O}_7 = \frac{52.8\, \text{g}}{201.22\, \text{g/mol}} \approx 0.262\, \text{mol}.\]

Step 4: Convert Water Mass to Moles

Calculate the number of moles of water using its molar mass, 18.02 g/mol:\[\text{moles of H}_2\text{O} = \frac{47.2\, \text{g}}{18.02\, \text{g/mol}} \approx 2.62\, \text{mol}.\]

Step 5: Determine Mole Ratio

Divide the moles of water by the moles of sodium tetraborate to find the mole ratio:\[\text{Mole ratio} = \frac{2.62\, \text{mol H}_2\text{O}}{0.262\, \text{mol Na}_2\text{B}_4\text{O}_7} \approx 10.\]

Step 6: Write the Hydrate Formula

The mole ratio indicates there are 10 water molecules per formula unit of sodium tetraborate. Therefore, the formula is \(\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}\).

Step 7: Name the Hydrate

The name of the compound, based on its hydrate formula, is sodium tetraborate decahydrate.

Key Concepts

Chemical Composition

Chemical composition refers to the elements and compounds that constitute a substance. In hydrate chemistry, understanding the chemical composition is critical. A hydrate is a compound that incorporates water molecules into its crystal structure. For example, the exercise involves hydrated sodium tetraborate, which is made up of sodium tetraborate and water. The percentage composition is given: 52.8% sodium tetraborate and 47.2% water.
By understanding chemical composition, you can start determining the proportions of each component within the compound. This helps in identifying the formula of the hydrate. Chemical analysis, like the one in the exercise, can inform you about what you're dealing with by showing you the exact percentage of each element or compound present. This information is crucial for further calculations.

Mole Ratio

The mole ratio is a vital concept when dealing with chemical formulas, especially hydrates. It involves comparing the number of moles of two substances within a compound. In the context of this exercise, you need to compare the moles of water to the moles of sodium tetraborate.
To find these numbers, you first convert the mass of each component to moles using their respective molar masses. For sodium tetraborate ( ext{Na}_2 ext{B}_4 ext{O}_7), the molar mass is 201.22 g/mol, and for water ( ext{H}_2 ext{O}), it's 18.02 g/mol. Calculating the moles helps determine how many molecules of water exist per molecule of the hydrate's compound.

• Use the formula: \[\text{Mole Ratio} = \frac{\text{moles of } H_2O}{\text{moles of } Na_2B_4O_7}\]

This yields a crucial number that directly informs the hydrate formula: the hydrate's water content.

Hydrate Formula

The hydrate formula represents the precise number of water molecules chemically bound within a compound's structure. After calculating the mole ratio in the exercise, you discovered there are 10 water molecules per formula unit of sodium tetraborate. This gives you the hydrate formula \[\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}\]This formula succinctly tells you that for every unit of sodium tetraborate, ten water molecules are included as part of its crystalline structure.
To arrive at this formula, understand the correlation that the mole ratio determines.

• Each number in a hydrate's formula corresponds to the real ratio of the different components found in the compound through experimental calculation.

Comprehending the formula allows you not only to define the chemical makeup but also to predict how it behaves chemically and physically.

Chemical Nomenclature

Nomenclature is the systematic approach to naming chemical compounds. In hydrate chemistry, this involves linking the name of the main compound with the prefix that reflects the number of water molecules. This process allows you to communicate the composition and proportion of the compound clearly.
In the given exercise, once the formula is established as \[\mathrm{Na}_2 \mathrm{B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O}\]you name the compound as sodium tetraborate decahydrate. The prefix "deca-" indicates the presence of ten water molecules.

• Some common hydrate prefixes include:
  • "Mono-" for one
  • "Di-" for two
  • "Tri-" for three
  • "Tetra-" for four
• Understanding chemical nomenclature assists in effectively communicating chemical information.

Each name reflects the chemical identity and composition, aiding in effective communication in scientific contexts.