Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH (188.35 g/mol) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{g} / \mathrm{mol})\) d. \(\mathrm{SN}(184.32 \mathrm{g} / \mathrm{mol})\)

Short Answer

Expert verified
The molecular formulas for the given compounds are: a. \(S_4N_4H_4\) b. \(Co_2C_8O_8\) c. \(N_3P_3Cl_6\) d. \(S_4N_4\)

Step by step solution

01

1. Calculate the molar mass of the empirical formula given.

First, we need to find the molar mass of the empirical formula for each of the compounds. To do this, we can use the periodic table to find the atomic mass of each element and then multiply it by the number of atoms present in the empirical formula.
02

a. SNH

Using the periodic table: Sulfur: 32.06 g/mol Nitrogen: 14.01 g/mol Hydrogen: 1.008 g/mol Molar mass of SNH = 32.06 + 14.01 + 1.008 = 47.078 g/mol
03

b. CoC4O4

Using the periodic table: Cobalt: 58.93 g/mol Carbon: 12.01 g/mol Oxygen: 16.00 g/mol Molar mass of CoC4O4 = 58.93 + (4 × 12.01) + (4 × 16.00) = 170.97 g/mol
04

c. NPCl2

Using the periodic table: Nitrogen: 14.01 g/mol Phosphorus: 30.97 g/mol Chlorine: 35.45 g/mol Molar mass of NPCl2 = 14.01 + 30.97 + (2 × 35.45) = 115.87 g/mol
05

d. SN

Using the periodic table: Sulfur: 32.06 g/mol Nitrogen: 14.01 g/mol Molar mass of SN = 32.06 + 14.01 = 46.07 g/mol
06

2. Find the ratio between the given molar mass and the molar mass of the empirical formulas.

Now, we divide the given molar mass by the molar mass of the empirical formula we calculated in step 1. This will give us the ratio of the molecular formula to the empirical formula.
07

a. Ratio for SNH

Ratio = (188.35 g/mol) / (47.078 g/mol) ≈ 4
08

b. Ratio for CoC4O4

Ratio = (341.94 g/mol) / (170.97 g/mol) ≈ 2
09

c. Ratio for NPCl2

Ratio = (347.64 g/mol) / (115.87 g/mol) ≈ 3
10

d. Ratio for SN

Ratio = (184.32 g/mol) / (46.07 g/mol) ≈ 4
11

3. Determine the molecular formula by multiplying the empirical formula by the ratio.

Using the ratio obtained, multiply the empirical formula by the ratio to find the molecular formula of the compounds.
12

a. Molecular formula for SNH

Molecular formula: SNH × 4 = S4N4H4
13

b. Molecular formula for CoC4O4

Molecular formula: CoC4O4 × 2 = Co2C8O8
14

c. Molecular formula for NPCl2

Molecular formula: NPCl2 × 3 = N3P3Cl6
15

d. Molecular formula for SN

Molecular formula: SN × 4 = S4N4 So the molecular formulas for the given compounds are: a. S4N4H4 b. Co2C8O8 c. N3P3Cl6 d. S4N4

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Key Concepts

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

Empirical Formula
Understanding the empirical formula is the foundation of many chemistry problems. An empirical formula represents the simplest whole-number ratio of the elements within a compound. It doesn't provide the exact number of atoms, but rather the relative number of atoms of each element present. For instance, the empirical formula for glucose is CH2O, which indicates that the ratio of carbon to hydrogen to oxygen is 1:2:1. However, glucose's molecular formula is C6H12O6, revealing that the substance is comprised of six times the number of atoms represented in the empirical formula.

To determine an empirical formula, one must find the lowest common denominator that can be used to express the ratio of elements in a compound as whole numbers. This often involves the use of molar mass calculations and a deep understanding of the chemical composition of the substance.
Molar Mass Calculation
A cornerstone of solving chemistry exercises is the ability to calculate molar masses. The molar mass of a substance corresponds to the weight of one mole of that substance, typically expressed in grams per mole (g/mol). One mole of a substance contains Avogadro's number (approximately 6.022 x 1023) of molecules or atoms.

To calculate the molar mass of a compound, one should add the molar masses of all individual elements present, weighted by their respective number of atoms in the formula. Let's take water (H2O) as an example: it has two hydrogen atoms and one oxygen atom. The molar mass of a hydrogen atom is approximately 1.008 g/mol, and for oxygen, it is about 16.00 g/mol. Thus, the molar mass of water equals (2 x 1.008) + (1 x 16.00) = 18.016 g/mol.

Practical Calculation

In the provided exercise, the molar mass of the empirical formula SNH was calculated by adding the molar masses of sulfur (32.06 g/mol), nitrogen (14.01 g/mol), and hydrogen (1.008 g/mol), leading to the sum of 47.078 g/mol. Such calculations are essential when converting between moles and grams, a common requirement in quantitative chemistry.
Chemical Composition
The chemical composition of a molecule is represented by its molecular formula, which provides information on the exact number of each type of atom present in the molecule. This distinguishes it from the empirical formula, which only reflects the simplest ratio. Chemical composition is central in learning about the characteristics and reactions of different substances.

Using the information about the molar mass and the empirical formula, one can deduce the molecular formula of a compound, which follows the steps highlighted in the exercise. It involves determining the ratio of the molar mass given to the molar mass of the empirical formula. This ratio is then used to multiply the subscripts in the empirical formula to obtain the molecular formula with actual counts of atoms. For example, if the empirical formula is CH2O and the molar mass indicates that the actual substance is eight times the mass of the empirical formula, it would imply the molecular formula is C8H16O8.

Application to Exercise

In the exercise, by calculating the ratio and multiplying it by the empirical formula, we find the molecular formula for each compound. The molecular formula not only gives us a clearer picture of the makeup of a compound but is also invaluable for stoichiometric calculations and understanding the physical and chemical properties of the substance.

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Most popular questions from this chapter

There are two binary compounds of mercury and oxygen. Heating either of them results in the decomposition of the compound, with oxygen gas escaping into the atmosphere while leaving a residue of pure mercury. Heating 0.6498 g of one of the compounds leaves a residue of 0.6018 g. Heating 0.4172 g of the other compound results in a mass loss of 0.016 g. Determine the empirical formula of each compound.

Acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is the starting material for many synthetic carpets and fabrics. It is produced by the following reaction. $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If \(15.0 \mathrm{g}\) \(\mathrm{C}_{3} \mathrm{H}_{6}, 10.0 \mathrm{g} \mathrm{O}_{2},\) and \(5.00 \mathrm{g}\) \(\mathrm{NH}_{3}\) are reacted, what mass of acrylonitrile can be produced, assuming \(100 \%\) yield?

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$\begin{array}{c}4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\\3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)\end{array}$$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{kg}\) \(\mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction, and assume that the NO produced in the third step is not recycled.

The most common form of nylon (nylon-6) is \(63.68 \%\) carbon, \(12.38 \%\) nitrogen, \(9.80 \%\) hydrogen, and \(14.14 \%\) oxygen. Calculate the empirical formula for nylon-6.

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced \(0.213 \mathrm{g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that 0.103 g of the compound produces \(0.0230 \mathrm{g} \mathrm{NH}_{3}\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

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