Problem 1 - Chemical Principles in the Laboratory

Question

A student is given a sample of a pink manganese(II) chloride hydrate. She weighs the sample in a dry, covered crucible and obtains a mass of \(26.742 \mathrm{g}\) for the crucible, cover, and sample. Earlier she had found that the crucible and cover weighed 23.599 g. She then heats the crucible to drive off the water of hydration, kceping the crucible at red heat for about 10 minutes with the cover slightly ajar. She then Iets the crucible cool and finds it has a lower mass; the crucible, cover and contents then weigh \(25.598 \mathrm{g}\). In the process the sample was converted to off-white anhydrous \(\mathrm{MnCl}_{2}\) a. What was the mass of the hydrate sample? _______ \(-g\) hydratc. b. What is the mass of the anhydrous \(\mathrm{MnCl}_{2} ?\) ______ \(\mathrm{g} \mathrm{MnCl}_{2}\). c. How much water was driven off? _______ \(-g \mathbf{H}_{2} \mathbf{O}\). What is the percent by mass of water in the hydrate? \% water \(=\frac{\text { mass of water in sample }}{\text { mass of hydrate sample }} \times 100 \%\) ________ \(\pi_{\max } \mathbf{H}_{2} \mathbf{O}\). e. How many grams of water would there be in \(100.0 \mathrm{g}\) hydrate? How many moles? ________ \(g H_{2} O\) ________ moles \(\mathrm{H}_{2} \mathrm{O}\). f. How many grams of \(\mathrm{MnCl}_{2}\) are there in \(100.0 \mathrm{g}\) hydrate? How many moles? (What percentage of the hydrate is \(\mathrm{MnCl}_{2}\) ? Convert the mass of \(\mathrm{MnC}_{2}\) to moles. The molar mass of \(\mathrm{MnCl}_{2}\) is \(125.85 \mathrm{g} / \mathrm{mol}\).) _______ \(g M n C l_{2}\) ______moles \(\mathrm{MnCl}_{2}\). g. How many moles of water are present per mole \(\mathrm{MnCl}_{2} ?\) |||||||||______ . h. What is the formula of the hydrate? ______ .

Step-by-Step Solution

Answer

3.143 g; 1.999 g; 1.144 g; 36.42%; 36.42 g, 2.022 moles; 63.58 g, 0.505 moles; 4.00; MnCl2·4H2O.

Step 1: Calculate Mass of Hydrate Sample

Subtract the mass of the crucible and cover only from the mass of the crucible, cover, and hydrate sample. \[\text{Mass of hydrate sample} = 26.742 \text{ g} - 23.599 \text{ g} = 3.143 \text{ g}\]

Step 2: Calculate Mass of Anhydrous MnCl2

Subtract the mass of the crucible and cover only from the mass of the crucible, cover, and anhydrous MnCl2. \[\text{Mass of anhydrous } \text{MnCl}_2 = 25.598 \text{ g} - 23.599 \text{ g} = 1.999 \text{ g}\]

Step 3: Calculate Mass of Water Driven Off

Subtract the mass of anhydrous MnCl2 from the mass of the hydrate sample.\[\text{Mass of water driven off} = 3.143 \text{ g} - 1.999 \text{ g} = 1.144 \text{ g}\]

Step 4: Calculate Percent by Mass of Water in Hydrate

Use the formula for percent by mass to find the percentage of water in the hydrate: \[\% \text{ water} = \left( \frac{1.144 \text{ g}}{3.143 \text{ g}} \right) \times 100\% = 36.42\%\]

Step 5: Calculate Grams and Moles of Water in 100 g Hydrate

For 100 grams of hydrate, use the percentage of water found in Step 4.\[\text{Grams of water} = 100 \text{ g} \times \frac{36.42}{100} = 36.42 \text{ g}\]Convert this to moles of water using the molar mass of water, which is 18.015 g/mol:\[\text{Moles of water} = \frac{36.42 \text{ g}}{18.015 \text{ g/mol}} = 2.022 \text{ moles}\]

Step 6: Calculate Grams and Moles of MnCl2 in 100 g Hydrate

Subtract percent water from 100 percent to get the percent of MnCl2:\[100\% - 36.42\% = 63.58\%\]Calculate the grams of MnCl2 in 100 g of hydrate:\[\text{Grams of MnCl}_2 = 100 \text{ g} \times \frac{63.58}{100} = 63.58 \text{ g}\]Convert this to moles of MnCl2 using its molar mass (125.85 g/mol):\[\text{Moles of MnCl}_2 = \frac{63.58 \text{ g}}{125.85 \text{ g/mol}} = 0.505 \text{ moles}\]

Step 7: Determine Moles of Water Per Mole of MnCl2

Divide the moles of water by the moles of MnCl2 to find the ratio:\[\text{Moles of water per mole of MnCl}_2 = \frac{2.022 \text{ moles}}{0.505 \text{ moles}} = 4.00\]

Step 8: Determine Formula of the Hydrate

Since there are approximately 4 moles of water for every mole of MnCl2, the formula of the hydrate is:\[\text{MnCl}_2 \cdot 4\text{H}_2\text{O}\]

Key Concepts

Percent Composition

Understanding percent composition is key in analyzing hydrates such as manganese(II) chloride hydrate. Percent composition refers to the percentage by mass of each component in a compound. In our example, we calculate the percent by mass of water in the hydrate using this formula \[\% \text{ water} = \left( \frac{\text{mass of water in sample}}{\text{mass of hydrate sample}} \right) \times 100\%\]To find the percent of water, we subtracted the mass of the anhydrous compound from the hydrated compound to get the water mass. Then, we divide this by the total mass of the hydrate and multiply by 100 to convert it into percentage. This calculation is key for determining how much of the sample's mass is due to water.

Percent composition helps to understand the structure of compounds, which is essential in chemistry. It provides insights into the proportion of elements present, which is vital for industrial applications like manufacturing medicines and materials. Knowing percent composition allows us to adjust formulas to obtain the desired compound with precise features.

Molar Mass Calculation

Knowing molar masses is crucial when dealing with chemical formulas and reactions. The molar mass is the mass of one mole of a given substance (molecule). It is typically expressed in grams per mole (g/mol).

For manganese(II) chloride (\(\text{MnCl}_2\)), the molar mass is calculated by adding the molar masses of the individual elements:

• Manganese (Mn) with a molar mass of about 54.94 g/mol.
• Chlorine (Cl) with a molar mass of about 35.45 g/mol.

Hence, the molar mass of \(\text{MnCl}_2 = 54.94 + 2(35.45) = 125.85 \text{ g/mol}\)
Using molar mass, we converted grams to moles in our given compound. For example, in the exercise, it helps in determining how many moles of water are associated with manganese chloride in 100 g of the hydrate. This concept is fundamental in chemistry, as it links the mass of a substance to the amount of substance. By understanding molar masses, you can predict the quantitative outcomes in chemical reactions and analyze the properties and compositions of chemical compounds accurately.

Chemical Formula Determination

Determining the chemical formula of a hydrate involves identifying the ratio of water molecules bound to each formula unit of the compound. In the problem, we calculated the moles of water and manganese chloride present in 100 g of the hydrate.

We found that 36.42 g of water equals about 2.022 moles (using water's molar mass of 18.015 g/mol), and 63.58 g of \(\text{MnCl}_2\) equals 0.505 moles (using its molar mass of 125.85 g/mol).The ratio of moles of water to moles of \(\text{MnCl}_2\) is found by dividing the number of moles of water by the number of moles of manganese chloride:\[\frac{2.022 \text{ moles of } \text{H}_2\text{O}}{0.505 \text{ moles of } \text{MnCl}_2} \approx 4\]This tells us that for every mole of \(\text{MnCl}_2\), there are approximately 4 moles of water. Thus, the chemical formula for the hydrate is identified as \(\text{MnCl}_2 \cdot 4\text{H}_2\text{O}\).

Determining the chemical formula is vital because it reveals the exact structure and composition of the compound. This knowledge is critical for predicting reaction behavior, understanding physical properties, and developing various industrial and pharmaceutical applications.