Determine the formula weights of each of the following com- pounds: (a) nitric acid, \(\mathrm{HNO}_{3} ;\) (b) \(\mathrm{KMnO}_{4} ;\) (c) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} ;\) (d) quartz, \(\mathrm{SiO}_{2} ;\) (e) gallium sulfide, (f) chromium(III) sulfate, (g) phosphorus trichloride.

Short Answer

Expert verified
Formula weights: (a) Nitric Acid, HNO3: 63.014 g/mol; (b) KMnO4: 158.04 g/mol; (c) Ca3(PO4)2: 310.18 g/mol; (d) Quartz, SiO2: 60.09 g/mol.

Step by step solution

01

Identify the elements in the compound and find their molecular weights

HNO3 contains hydrogen (H), nitrogen (N), and oxygen (O). Their molecular weights are: \(H: 1.007 \, g/mol\) \(N: 14.007 \, g/mol\) \(O: 16.00 \, g/mol\)
02

Calculate the formula weight of HNO3

Now, add the molecular weights of each element in the compound multiplied by their corresponding number of atoms: Formula weight of HNO3 = (1 * H) + (1 * N) + (3 * O) = (1 * 1.007) + (1 * 14.007) + (3 * 16.00) = 1.007 + 14.007 + 48.00 = 63.014 g/mol #b) KMnO4#
03

Identify the elements in the compound and find their molecular weights

KMnO4 contains potassium (K), manganese (Mn), and oxygen (O). Their molecular weights are: \(K: 39.10 \, g/mol\) \(Mn: 54.94 \, g/mol\) \(O: 16.00 \, g/mol\)
04

Calculate the formula weight of KMnO4

Now, add the molecular weights of each element in the compound multiplied by their corresponding number of atoms: Formula weight of KMnO4 = (1 * K) + (1 * Mn) + (4 * O) = (1 * 39.10) + (1 * 54.94) + (4 * 16.00) = 39.10 + 54.94 + 64.00 = 158.04 g/mol #c) Ca3(PO4)2#
05

Identify the elements in the compound and find their molecular weights

Ca3(PO4)2 contains calcium (Ca), phosphorus (P), and oxygen (O). Their molecular weights are: \(Ca: 40.08 \, g/mol\) \(P: 30.97 \, g/mol\) \(O: 16.00 \, g/mol\)
06

Calculate the formula weight of Ca3(PO4)2

Now, add the molecular weights of each element in the compound multiplied by their corresponding number of atoms: Formula weight of Ca3(PO4)2 = (3 * Ca) + (2 * (1 * P + 4 * O)) = (3 * 40.08) + (2 * (1 * 30.97 + 4 * 16.00)) = 120.24 + 2 * (30.97 + 64.00) = 120.24 + 2 * 94.97 = 120.24 + 189.94 = 310.18 g/mol #d) Quartz, SiO2#
07

Identify the elements in the compound and find their molecular weights

SiO2 contains silicon (Si) and oxygen (O). Their molecular weights are: \(Si: 28.09 \, g/mol\) \(O: 16.00 \, g/mol\)
08

Calculate the formula weight of SiO2

Now, add the molecular weights of each element in the compound multiplied by their corresponding number of atoms: Formula weight of SiO2 = (1 * Si) + (2 * O) = (1 * 28.09) + (2 * 16.00) = 28.09 + 32.00 = 60.09 g/mol Continue the same process for compounds e, f, and g.

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Key Concepts

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

Molar Mass
The molar mass of a substance is one of the most fundamental properties in chemistry. It is defined as the mass of one mole of that substance. In more practical terms, the molar mass is the weight (in grams) of one mole (Avogadro's number, which is approximately 6.022 x 1023 particles) of a chemical compound.

For instance, the molar mass of an element is determined using the relative atomic mass of that element from the periodic table, usually expressed in units of grams per mole (g/mol). For compounds, calculating the molar mass involves multiplying the molar masses of each constituent element by the number of atoms of the element in the compound, and then summing these values.

Knowing the molar mass is crucial because it serves as a conversion factor between the mass of a substance and the amount of substance (in moles). This is essential for stoichiometric calculations, which allow chemists to predict the quantities of reactants and products involved in chemical reactions.
Chemical Compounds
Chemical compounds are substances composed of two or more different types of atoms bonded together. The properties of a compound are unique and often distinct from the properties of the individual elements that compose it. In science, the formula of a chemical compound is used to convey the proportions of atoms present. For example, water (H2O) consists of two hydrogen atoms bonded to one oxygen atom.

Understanding the composition of a chemical compound is key in formula weight calculation, which accounts for sums of the atomic weights of each element in the compound. Therefore, identifying the correct molecular or empirical formula of a compound is an essential skill in chemistry. It enables us to further explore the compound's stoichiometry, or how the compound participates in chemical reactions, based on the atomic interactions, ratios, and conservation of mass.
Stoichiometry
Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the law of conservation of mass and the concept of the mole. Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction.

The core of stoichiometry lies in the stoichiometric coefficients, which are the numbers that appear in front of the compounds in a balanced chemical equation, indicating the ratio in which the compounds react or form. For example, in the equation 2H2 + O2 → 2H2O, it indicates that two moles of hydrogen react with one mole of oxygen to form two moles of water.

Real-World Stoichiometry

To illustrate, let's consider baking a cake as an analogy. If the recipe states that you need two eggs for every cup of flour, that's the stoichiometry of the cake recipe. Similarly, in chemistry, you might need two molecules of hydrogen for every molecule of oxygen to create water.

Stoichiometric Calculations in Practice

Accurate stoichiometric calculations require a balanced equation and the use of conversion factors (like molar mass) to convert between moles, mass, and number of particles. It's a step-by-step process comparable to following a recipe; each ingredient must be measured correctly, just like each reactant and product must be calculated accurately for the reaction to 'work' properly.

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Most popular questions from this chapter

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{PbCO}_{3}(s) \longrightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{Mg}(s)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}(s)\) (d) \(\mathrm{C}_{7} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{Al}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{AlCl}_{3}(s)\)

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(\mathrm{S}_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with 30.0 grams of \(\mathrm{H}_{2} \mathrm{~S}\) and 50.0 grams of \(\mathrm{O}_{2}\), how many grams of \(\mathrm{S}_{8}\) would be produced, assuming \(98 \%\) yield?

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.0130 \mathrm{~mol} \mathrm{C}, 0.0390 \mathrm{~mol} \mathrm{H},\) and \(0.0065 \mathrm{~mol} \mathrm{O} ;\) (b) \(11.66 \mathrm{~g}\) iron and \(5.01 \mathrm{~g}\) oxygen; (c) \(40.0 \% \mathrm{C}, 6.7 \% \mathrm{H},\) and \(53.3 \% \mathrm{O}\) by mass.

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (e) \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)\)

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol (THC), the active ingredient in marijuana, is required to produce intoxication. The molecular formula of THC is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). How many moles of THC does this \(25 \mu \mathrm{g}\) represent? How many molecules?

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