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Chapter 3: Problem 108

When \(0.273 \mathrm{~g}\) of \(\mathrm{Mg}\) is heated strongly in a nitrogen \(\left(\mathrm{N}_{2}\right)\) atmosphere, a chemical reaction occurs. The product of the reaction weighs 0.378 g. Calculate the empirical formula of the compound containing \(\mathrm{Mg}\) and \(\mathrm{N}\). Name the compound.

Short Answer

Expert verified
The empirical formula is \(\text{Mg}_3\text{N}_2\), and the compound is Magnesium Nitride.

Step by step solution

01

Determine the Mass of Nitrogen

To find the mass of nitrogen that reacted, subtract the mass of magnesium from the total mass of the compound: \[ ext{Mass of N} = 0.378 ext{ g (product)} - 0.273 ext{ g (Mg)} = 0.105 ext{ g (N)} \]
02

Convert Masses to Moles

Next, convert the masses of Mg and N to moles by dividing by their respective molar masses. The molar mass of Mg is 24.305 g/mol and that of N is 14.007 g/mol. \[ ext{Moles of Mg} = \frac{0.273 ext{ g}}{24.305 ext{ g/mol}} = 0.01123 ext{ mol} \] \[ ext{Moles of N} = \frac{0.105 ext{ g}}{14.007 ext{ g/mol}} = 0.00750 ext{ mol} \]
03

Find the Simplest Mole Ratio

Determine the simplest whole-number mole ratio of Mg to N by dividing each value by the smallest number of moles calculated. \[ ext{Ratio of Mg} = \frac{0.01123}{0.00750} = 1.497 \approx 1.5 \] \[ ext{Ratio of N} = \frac{0.00750}{0.00750} = 1 \] Since we have a ratio of approximately 1.5 Mg to 1 N, we multiply both by 2 to get whole numbers: \[ ext{Mg}: 1.5 imes 2 = 3, ext{N}: 1 imes 2 = 2 \]
04

Write the Empirical Formula

The simplest whole-number ratio of Mg to N is 3:2. Thus, the empirical formula for the compound is \(\text{Mg}_3\text{N}_2 \).
05

Name the Compound

Based on the empirical formula \(\text{Mg}_3\text{N}_2\), the compound is named Magnesium Nitride.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula
The empirical formula is a fundamental concept in chemistry that gives the simplest whole-number ratio of elements in a compound. It is different from the molecular formula, which tells you the exact number of atoms of each element in a compound. To find the empirical formula, we start by determining the masses of each element in the compound and convert these masses to moles.
This process involves using the molar mass of each element, which serves as a conversion factor between grams and moles. Once we have the mole quantities, we find the ratio of the moles of each element by dividing by the smallest number of moles we calculated.
This gives a simplified ratio that usually needs adjustment to whole numbers, as chemical formulas are expressed in whole numbers. In the example with magnesium and nitrogen, the ratio of Mg to N calculated was 1.5 to 1. Multiplying both by 2 resulted in the empirical formula ext{Mg}_3 ext{N}_2, reflecting the simplest ratio of three magnesium atoms to two nitrogen atoms.
Magnesium Nitride
Magnesium nitride is a binary compound consisting of magnesium and nitrogen. In the chemical process described, when magnesium is heated in a nitrogen atmosphere, they react to form magnesium nitride. The chemical formula for magnesium nitride is ext{Mg}_3 ext{N}_2, indicating that every molecule of this compound contains three magnesium atoms bonded to two nitrogen atoms.
Magnesium nitride is significant due to its chemical properties, such as being a nitride-forming compound that can be used to synthesize other materials. When handling this compound, it鈥檚 essential to remember that it forms through a direct reaction between elemental magnesium and nitrogen gas.
Thus, knowing the empirical formula of this compound not only gives us insight into its composition but also into its chemical behavior and potential reactions.
Mole Ratio
The mole ratio is a crucial part of determining the empirical formula of a compound in chemical reactions. It refers to the proportion of moles of each element present in a compound. Understanding this ratio is important because it helps chemists deduce the simplest formula of a compound.
To find the mole ratio, the number of moles of each element is divided by the smallest number of moles in any of the elements present. This mathematical operation normalizes the ratios, making it possible to represent them in the simplest whole-number form.
In the case of magnesium nitride, once the number of moles of magnesium and nitrogen were determined, dividing both by the smaller number of moles gave a ratio of approximately 1.5:1 for Mg to N. Multiplying both by 2 provided a clean whole number ratio of 3:2, leading to the empirical formula of ext{Mg}_3 ext{N}_2. This systematic approach ensures that the empirical formula accurately reflects the compound's composition.

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Most popular questions from this chapter

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When \(X\) is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent \(\mathrm{Mn}\) and 28.0 percent \(\mathrm{O}\) is formed. (a) Determine the empirical formulas of \(X\) and Y. (b) Write a balanced equation for the conversion of \(X\) to \(Y\).

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ How much sulfur (in tons), present in the original materials, would result in that quantity of \(\mathrm{SO}_{2} ?\)

What are the empirical formulas of the compounds with the following compositions? (a) 2.1 percent \(\mathrm{H}\) 65.3 percent \(\mathrm{O}, 32.6\) percent \(\mathrm{S},\) (b) 20.2 percent \(\mathrm{Al}\) 79.8 percent \(\mathrm{Cl}\).

Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; \(\mathrm{H}: 6.21\) percent; \(\mathrm{S}: 39.5\) percent; \(\mathrm{O}:\) 9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is about \(162 \mathrm{~g}\) ?

The molar mass of caffeine is \(194.19 \mathrm{~g}\). Is the molecular formula of caffeine \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}\) or \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\)
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