Question

A student is given a sample of a pink manganese(II) sulfate 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. Earlicr 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 ten minutes with the cover slightly ajar. She then lets the crucible cool and finds it has a lower mass; the crucible, cover, and contents then weigh \(26.406 \mathrm{g}\). In the process the sample was converted to off-white anhydrous \(\mathrm{MnSO}_{4}\) a. What was the mass of the hydrate sample? g hydrate What is the mass of the anhydrous \(\mathrm{MnSO}_{4} ?\) $$\mathrm{g} \mathrm{MnSO}_{4}$$ c. How much water was driven off? $$g H_{2} O$$ c. How much water was driven off? $$\mathrm{g} \mathrm{H}_{2} \mathrm{O}$$ d. What is the percent by mass of water in the hydrate? \(\%_{\text {mass }}\) water \(=\frac{\text { mass of water in sample }}{\text { mass of hydrate sample }} \times 100 \%\) $$\%_{\text {mass }} \mathrm{H}_{2} \mathrm{O}$$ e. How many grams of water would there be in \(100.0 \mathrm{g}\) hydrate? How many moles? $$\begin{aligned} &g H_{2} O_{i}\\\ &\text { moles } \mathrm{H}_{2} \mathrm{O} \end{aligned}$$ f. How many grams of MnSO_are there in \(100.0 \mathrm{g}\) hydrate? How many moles? (What percentage of the mass of the hydrate is \(\mathrm{MnSO}_{4}\) ? Convert the mass of \(\mathrm{MnSO}_{4}\) to moles. The molar mass of \(\mathrm{MnSO}_{4}\) is 151.00 g/mol.) \(\mathrm{g} \mathrm{MnSO}_{4}:\) moles \(\mathrm{MnSO}_{4}\) g. How many moles of water are present per mole of \(\mathrm{MnSO}_{4} ?\) h. What is the formula of the hydrate?

Short answer

The formula of the manganese(II) sulfate hydrate is \(MnSO_4 \cdot H_2O\).

Step-by-step solution

Calculate the mass of the hydrate sample

The mass of the crucible, cover and the hydrate sample is 26.742 g and the mass of the crucible and cover alone is 23.599 g. Therefore the mass of the hydrate sample can be found by subtracting the mass of the crucible and cover from the total mass, which is \(26.742 \, \text{g}- 23.599 \, \text{g} = 3.143 \, \text{g} \).

Establish the mass of the anhydrous MnSO4

After heating, the mass of the crucible, cover and the anhydrous sample is 26.406 g. The mass of the manganese sulfate can be determined by subtracting the mass of the crucible and cover from the mass after heating, which is \(26.406 \, \text{g}- 23.599 \, \text{g} = 2.807 \, \text{g} \).

Determine the amount of water driven off

The amount of water displaced due to the heating can be found by subtracting the mass of the anhydrous manganese sulfate from the initial hydrate, that is, \(3.143 \, \text{g} - 2.807 \, \text{g} = 0.336 \, \text{g} \).

Find the percent by mass of water in the hydrate

This fraction is determined by the ratio of the mass of water to the total mass of the hydrate, multiplied by 100%. Thus the percent by mass is \(\big(\frac{0.336 \, \text{g}}{3.143 \, \text{g}}\big) \times 100 \% = 10.69 \% \).

Calculate the grams of water and manganese sulfate in 100g hydrate

Using the percent by mass, it is easily seen that in 100g of the hydrate there would be 10.69 g of water and 89.31 g of manganese sulfate MnSO4.

Determine the number of moles

The molar mass of water H2O is 18.015 g/mol, and that of MnSO4 is 151.00 g/mol. The number of moles of each can be given by the ratio of the mass to the molar mass. There are \(\frac{10.69 \, \text{g}}{18.015 \, g/mol} = 0.593 \, \text{moles}\) of water and \(\frac{89.31 \, \text{g}}{151.00 \, g/mol} = 0.591 \, \text{moles}\) of MnSO4.

Establish the formula of the hydrate

This ratio gives the number of moles of water per mole of MnSO4, which would be \(0.593:0.591\), almost 1:1. The formula of the hydrate is therefore \(MnSO4 \cdot H2O\).

Key concepts

Stoichiometry

Stoichiometry is like the math of chemistry, a fundamental concept used to calculate the relationships between reactants and products in chemical reactions. It involves using balanced chemical equations to determine the proportions of substances needed or produced. In the context of lab experiments, stoichiometry can help predict how much product will form from given reactants. This is crucial when trying to understand and quantify reactions.
For example, when water evaporates from a hydrated compound, stoichiometry helps us calculate the precise amounts of each component left behind. In simple terms, by knowing the quantities of water and manganese sulfate in our original hydrate, we can use stoichiometry to determine the exact reaction and formulate the compound.

Hydrate Analysis

Hydrate analysis examines compounds that include water molecules as part of their structure. When you come across a compound labeled as "hydrate," it means water is chemically bonded within the crystal structure. During heating, this water is removed, resulting in an anhydrous or water-free form. This change can be measured to determine how much water was initially present.
In the given experiment, the transition from pink manganese(II) sulfate hydrate to an off-white anhydrous manganese sulfate demonstrates hydrate analysis. By knowing the mass before and after heating, it's possible to quantify the amount of water lost. This process helps in identifying the water of crystallization and plays a key role in deriving chemical formulas for hydrates.

Mass Percent Calculation

Mass percent is a way of describing how much of a particular substance exists in a compound, expressed as a percentage of the total mass. To calculate, you divide the mass of the substance of interest by the total mass, then multiply by 100.
In the manganese sulfate hydrate example, the calculation involves finding the mass percent of water: \(\left(\frac{0.336 \text{ g}}{3.143 \text{ g}}\right) \times 100\% = 10.69\%\). This tells us that water makes up 10.69% of the total hydrate mass. Such calculations are crucial in chemistry labs for determining compositions and adjusting formulas. By understanding mass percent, one can infer how molecular water contributes to the overall structure of a hydrate.

Moles and Molar Mass

Moles and molar mass are core concepts in understanding chemical quantities. A mole is a unit of measurement that represents \(6.022 \times 10 ^{23}\) particles, atoms, or molecules of a substance—commonly known as Avogadro's number. Molar mass, on the other hand, is the mass of one mole of a substance and is expressed in grams per mole (g/mol).
The manganese sulfate hydrate experiment provides an excellent example. For water, the molar mass is 18.015 g/mol, and for anhydrous manganese sulfate (MnSO₄), it's 151.00 g/mol. These values are used to convert mass into moles, which then helps balance the chemical equation and assists in determining the formula of the hydrate. By utilizing moles and molar mass, you can convert between weight and the number of molecules, providing insights into chemical reactions.

Chemical Formulas

Chemical formulas represent the types and numbers of atoms in a molecule. For hydrates, they include information about water molecules integrated within the compound, denoted as \(\cdot H_2O\). This conveys both chemical and physical information, indicating how substances interact.
In the experiment, we derive the formula for manganese(II) sulfate hydrate based on the stoichiometric analysis and hydrate water content calculations. Initially, we suspect a general form \(MnSO_4 \cdot xH_2O\), where \(x\) represents the number of water molecules. Analysis based on mass and moles can ascertain this factor, allowing precise depiction of the chemical structure. Knowing the precise formula enables chemists to predict and understand the properties and reactions of the compound within various contexts.