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Determine the formula weights of each of the following compounds: (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
The formula weights for the compounds are as follows: (a) Nitric Acid (HNO₃): \(1\times\text{H} + 1\times\text{N} + 3\times\text{O}\) (b) Potassium Permanganate (KMnO₄): \(1\times\text{K} + 1\times\text{Mn} + 4\times\text{O}\) (c) Calcium Phosphate (Ca₃(PO₄)₂): \(3\times\text{Ca} + 2\times(1\times\text{P} + 4\times\text{O})\) (d) Quartz (SiO₂): \(1\times\text{Si} + 2\times\text{O}\) (e) Gallium Sulfide (Ga₂S₃): \(2\times\text{Ga} + 3\times\text{S}\) (f) Chromium(III) Sulfate (Cr₂(SO₄)₃): \(2\times\text{Cr} + 3\times(1\times\text{S} + 4\times\text{O})\) (g) Phosphorus Trichloride (PCl₃): \(1\times\text{P} + 3\times\text{Cl}\)

Step by step solution

01

(a) Formula weight of Nitric Acid (HNO₃)

1. Find the atomic weights of hydrogen (H), nitrogen (N), and oxygen (O) from the periodic table. 2. Multiply the atomic weight of each element by the number of atoms it has in the chemical formula. 3. Add the weighted elements to get the formula weight. \[1 \times \text{H} + 1 \times \text{N} + 3 \times \text{O} = \text{(atomic weight of H)} + \text{(atomic weight of N)} + 3\times\text{(atomic weight of O)}\]
02

(b) Formula weight of Potassium Permanganate (KMnO₄)

1. Find the atomic weights of potassium (K), manganese (Mn), and oxygen (O) from the periodic table. 2. Add the atomic weights for each element in the chemical formula. \[1 \times \text{K} + 1 \times \text{Mn} + 4 \times \text{O} = \text{(atomic weight of K)} + \text{(atomic weight of Mn)} + 4\times\text{(atomic weight of O)}\]
03

(c) Formula weight of Calcium Phosphate (Ca₃(PO₄)₂)

1. Find the atomic weights of calcium (Ca), phosphorus (P), and oxygen (O) from the periodic table. 2. Add the atomic weights for each element taking into account the subscripts in the chemical formula. \[3\times\text{Ca} + 2\times(1\times\text{P} + 4\times\text{O}) = 3\times\text{(atomic weight of Ca)} + 2\times(\text{(atomic weight of P)} + 4\times\text{(atomic weight of O)})\]
04

(d) Formula weight of Quartz (SiO₂)

1. Find the atomic weights of silicon (Si) and oxygen (O) from the periodic table. 2. Add the atomic weights for each element in the chemical formula. \[1 \times \text{Si} + 2 \times \text{O} = \text{(atomic weight of Si)} + 2\times\text{(atomic weight of O)}\]
05

(e) Formula weight of Gallium Sulfide (Ga₂S₃)

1. Find the atomic weights of gallium (Ga) and sulfur (S) from the periodic table. 2. Add the atomic weights for each element in the chemical formula. \[2 \times \text{Ga} + 3 \times \text{S} = 2\times\text{(atomic weight of Ga)} + 3\times\text{(atomic weight of S)}\]
06

(f) Formula weight of Chromium(III) Sulfate (Cr₂(SO₄)₃)

1. Find the atomic weights of chromium (Cr), sulfur (S), and oxygen (O) from the periodic table. 2. Add the atomic weights for each element taking into account the subscripts in the chemical formula. \(2\times\text{Cr} + 3\times(1\times\text{S} + 4\times\text{O}) = 2\times\text{(atomic weight of Cr)} + 3\times(\text{(atomic weight of S)} + 4\times\text{(atomic weight of O)})\)
07

(g) Formula weight of Phosphorus Trichloride (PCl₃)

1. Find the atomic weights of phosphorus (P) and chlorine (Cl) from the periodic table. 2. Add the atomic weights for each element in the chemical formula. \[1 \times \text{P} + 3 \times \text{Cl} = \text{(atomic weight of P)} + 3\times\text{(atomic weight of Cl)}\]

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Key Concepts

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

Atomic Weights
Understanding atomic weights is essential for anyone delving into the world of chemistry. Atomic weight, also referred to as atomic mass, is the weighted average mass of the atoms of an element based on the isotopes present in nature. It reflects the relative abundance of each isotope. These weights are not whole numbers because they account for the mix of isotopes and their respective masses.

For students and chemists alike, the atomic weight serves as a critical building block in calculations. It's the 'unit' used to quantify mass at the atomic and molecular level, which allows us to bridge the gap between the microscopic world of atoms and the macroscopic world we can measure in the lab. When it comes to formula weight calculation, it means multiplying the atomic weight of each element by the number of atoms of that element in the molecule and summing these values to represent the weight of the entire molecule.
Chemical Formulas
Chemical formulas represent the composition of a compound. They list the elements contained within the compound and indicate the number of atoms of each, using subscript numbers. For example, the formula for water, H2O, tells us that a water molecule consists of two hydrogen atoms bonded to one oxygen atom.

When calculating formula weights, the chemical formula is your roadmap. You need to read it carefully to determine the number of atoms of each element present. It is also crucial to take into account the subscripts, which indicate the quantity of atoms, and parentheses, which often multiply the number inside. This ensures the calculation is accurate and reflects the true makeup of the compound. Missing this step can lead to incorrect results, affecting the outcomes of stoichiometric calculations and other related tasks.
Periodic Table
The periodic table is not just a chart; it's the cheat sheet of chemistry. It organizes all known elements by increasing atomic number and groups them by shared chemical properties. The atomic weights needed for formula weight calculations are found here. For a student tackling these calculations, the periodic table is an invaluable reference.

The table provides more than just atomic weights. It also gives insight into the relationships between different elements and how they might interact with each other. Students can leverage this information to predict the behaviour of compounds and to help understand why elements combine in the ratios they do. Each element's position in the periodic table tells a story about its electrons, and this in turn explains how it forms compounds—which all circles back to understanding the balances seen in chemical formulas.
Stoichiometry
Stoichiometry is essentially the art of balancing recipes in chemistry. It takes into account the quantitative relationships that exist within a chemical reaction. These variable relationships between reactants and products illustrate the conservation of mass and the concept that atoms are neither created nor destroyed, simply rearranged.

In stoichiometric calculations, formula weights play a significant role as they allow chemists to convert between moles, which measure quantity, and grams, which measure weight. This conversion is pivotal for quantifying reactants and products in a reaction. Stoichiometry is all about proportion and balance, and without precise formula weight calculations—based on the accurate application of atomic weights and chemical formulas—this delicate balance would be impossible to achieve.

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Most popular questions from this chapter

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Serotonin is a compound that conducts nerve impulses in the brain. It contains \(68.2\) mass percent \(C, 6.86\) mass percent \(H\), \(15.9\) mass percent \(\mathrm{N}\), and \(9.08\) mass percent \(\mathrm{O}\). Its molar mass is \(176 \mathrm{~g} / \mathrm{mol}\). Determine its molecular formula.

Give the empirical formula of each of the following com\(\mathrm{H}\), 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.

(a) One molecule of the antibiotic penicillin \(\mathrm{G}\) has a mass of \(5.342 \times 10^{-21} \mathrm{~g}\). What is the molar mass of penicillin \(\mathrm{G}\) ? (b) Hemoglobin, the oxygen-carrying protein in red blood cells, has four iron atoms per molecule and contains \(0.340 \%\) iron by mass. Calculate the molar mass of hemoglobin.

A sample of the male sex hormone testosterone, \(\mathrm{C}_{19} \mathrm{H}_{28} \mathrm{O}_{2}\), contains \(3.88 \times 10^{21}\) hydrogen atoms. (a) How many atoms of carbon does it contain? (b) How many molecules of testosterone does it contain? (c) How many moles of testosterone does it contain? (d) What is the mass of this sample in grams?

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