Chapter 8: Problem 123
When \(2.004 \mathrm{g}\) of calcium is heated in pure nitrogen gas, the sample gains \(0.4670 \mathrm{g}\) of nitrogen. Calculate the empirical formula of the calcium nitride formed.
Short Answer
Expert verified
The empirical formula for the calcium nitride formed is Ca_3N_2.
Step by step solution
01
Convert mass to moles
First, we need to convert the grams of calcium and nitrogen to moles using their respective molar masses. The molar mass of calcium is 40.08 g/mol and the molar mass of nitrogen is 14.01 g/mol.
Moles of calcium = \(\frac{2.004 \text{ g}}{40.08 \text{ g/mol}} = 0.0500 \text{ mol}\)
Moles of nitrogen = \(\frac{0.4670 \text{ g}}{{14.01} \text{ g/mol}} = 0.03333 \text{ mol}\)
02
Determine the mole ratio
Next, we need to find the mole ratio of the calcium to nitrogen atoms in the compound.
To do this, we divide the moles of each element by the smaller amount of moles:
Mole ratio of calcium to nitrogen = \(\frac{0.0500 \text{ mol}}{0.03333 \text{ mol}} = 1.50\)
Since the mole ratio is 1.50, we can't simply round it to the nearest whole number as that would be 2, making the ratio 2:1. This would not be accurate, as it would mean that half the nitrogen atoms have disappeared. Instead, we should multiply both numbers by 2 to get the smallest whole number ratio:
Mole ratio of calcium to nitrogen = \(1.50 \times 2 : 1 \times 2 = 3 : 2\)
03
Write the empirical formula
Now that we have the mole ratio of calcium to nitrogen as 3:2, we can write the empirical formula:
The empirical formula for the calcium nitride formed is Ca_3N_2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mole Concept
Understanding the mole concept is fundamental in chemistry, as it forms the bridge between the atomic world and practical measurements in the laboratory. The mole, symbolized by 'mol,' is one of the seven base units in the International System of Units (SI) and is defined as the amount of a substance that contains as many entities (atoms, molecules, ions, etc.) as there are atoms in 12 grams of carbon-12.
When we deal with substances in the real world, we need to relate their masses in grams to the number of particles they have. One mole of any substance contains exactly Avogadro's number, or approximately 6.022\(\times\)10^23 entities, no matter what that substance is. For instance, one mole of calcium atoms contains 6.022\(\times\)10^23 calcium atoms.
In general, to convert from mass to moles, we use the formula:
\[\text{Moles} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}\]
Applying this formula to our exercise, we can understand that first, we must determine the number of moles of each element present by using their respective masses and molar masses.
When we deal with substances in the real world, we need to relate their masses in grams to the number of particles they have. One mole of any substance contains exactly Avogadro's number, or approximately 6.022\(\times\)10^23 entities, no matter what that substance is. For instance, one mole of calcium atoms contains 6.022\(\times\)10^23 calcium atoms.
In general, to convert from mass to moles, we use the formula:
\[\text{Moles} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}\]
Applying this formula to our exercise, we can understand that first, we must determine the number of moles of each element present by using their respective masses and molar masses.
Chemical Stoichiometry
Chemical stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculating the amounts of substances needed or produced based on balanced chemical equations and is rooted in the conservation of mass and the concept of the mole. Stoichiometry allows chemists to make predictions about the outcomes of reactions under specific conditions.
To perform stoichiometry calculations, we interpret chemical equations in terms of moles and use the mole ratio, which is derived from the coefficients of the balanced equation. In the context of finding empirical formulas, a key step is to determine the simplest whole number ratio of moles of each element in the compound. This involves comparing the moles of each element by dividing by the smallest number of moles among them.
For example, if the ratio is not an integer, as we see with the 1.5:1 ratio of calcium to nitrogen in our exercise, we multiply through to find the smallest whole numbers that represent this ratio accurately. This method ensures we do not lose or gain atoms, adhering to the law of conservation of mass.
To perform stoichiometry calculations, we interpret chemical equations in terms of moles and use the mole ratio, which is derived from the coefficients of the balanced equation. In the context of finding empirical formulas, a key step is to determine the simplest whole number ratio of moles of each element in the compound. This involves comparing the moles of each element by dividing by the smallest number of moles among them.
For example, if the ratio is not an integer, as we see with the 1.5:1 ratio of calcium to nitrogen in our exercise, we multiply through to find the smallest whole numbers that represent this ratio accurately. This method ensures we do not lose or gain atoms, adhering to the law of conservation of mass.
Molar Mass
Molar mass is a fundamental concept in chemistry and is defined as the mass of one mole of a given substance, expressed in grams per mole (g/mol). It is numerically equivalent to the atomic or molecular weight of the substance but used in a different context. When looking at the periodic table, the atomic weight of an element represents the molar mass for a single element. For compounds, the molar mass is the sum of the atomic weights of each atom in the molecular formula.
Knowing the molar mass allows us to convert between the mass of a substance and the number of moles using the formula mentioned previously. It serves as a conversion factor and plays a critical role in the calculation of empirical formulas. In the exercise at hand, the molar masses of calcium (40.08 g/mol) and nitrogen (14.01 g/mol) are used to determine how many moles of each element are present in the initial sample. The outcome of these calculations is critical for determining the correct empirical formula of the calcium nitride formed.
Knowing the molar mass allows us to convert between the mass of a substance and the number of moles using the formula mentioned previously. It serves as a conversion factor and plays a critical role in the calculation of empirical formulas. In the exercise at hand, the molar masses of calcium (40.08 g/mol) and nitrogen (14.01 g/mol) are used to determine how many moles of each element are present in the initial sample. The outcome of these calculations is critical for determining the correct empirical formula of the calcium nitride formed.