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Chapter 8: Problem 71

A compound has been analyzed and has been found to have the following composition: copper, \(66.75 \%\) phosphorus, \(10.84 \% ;\) oxygen, \(22.41 \% .\) Determine the empirical formula of the compound.

Short Answer

Expert verified
The empirical formula of the compound is Cu鈧働O鈧�.

Step by step solution

01

Convert percentages to grams

We are given the percentage composition of the compound: - Copper: 66.75% - Phosphorus: 10.84% - Oxygen: 22.41% To convert the percentages to grams, we will assume we have 100 grams of the compound. This means the compound contains: - 66.75 g of copper - 10.84 g of phosphorus - 22.41 g of oxygen
02

Convert grams to moles

The molar mass of each element is as follows: - Copper: 63.55 g/mol - Phosphorus: 30.97 g/mol - Oxygen: 16.00 g/mol To convert grams to moles, we will use the formula: Moles = (grams) / (molar mass) For each element: - Moles of copper = \( \frac{66.75}{63.55} \) = 1.05 mol - Moles of phosphorus = \( \frac{10.84}{30.97} \) = 0.35 mol - Moles of oxygen = \( \frac{22.41}{16.00} \) = 1.40 mol
03

Get the mole ratios

Divide all the moles by the smallest mole value: - Copper: \( \frac{1.05}{0.35} \) = 3 - Phosphorus: \( \frac{0.35}{0.35} \) = 1 - Oxygen: \( \frac{1.40}{0.35} \) = 4
04

Write the empirical formula

Using the whole number ratios from step 3, we can now write the empirical formula of the compound: Cu鈧働鈧丱鈧� Hence, the empirical formula of the compound is Cu鈧働O鈧�.

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Key Concepts

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

Understanding Mole Calculations
Mole calculations are essential for converting between the mass of a substance and the number of particles it contains. A mole is a specific quantity used in chemistry to express amounts of a chemical substance. One mole is equivalent to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles, be that atoms, molecules, or ions. We often use mole calculations to convert mass to moles by using the formula:\[\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\]
For example, in our compound, the percentages were first converted to grams by assuming a 100 g sample. Then, we used the molar mass of each element to find the number of moles. Knowing how to perform these calculations helps us determine the relationships and proportions between elements in a compound.
Using mole calculations allows chemists to deduce empirical formulas, which show the simplest ratio of elements within a compound. These formulas are essential for understanding the compound's composition and the proportions in which its atoms combine.
Exploring Percentage Composition
Percentage composition refers to the percentage by mass of each element in a compound. It is a crucial concept in chemistry because it helps identify the constituents of a compound. By knowing the percentage composition, chemists can infer possible empirical formulas and blend ratios for chemical reactions.
In practical terms, percentage composition is determined by dividing the mass of each element by the total molar mass of the compound and multiplying the result by 100. For our compound, we had three elements: copper, phosphorus, and oxygen. Assuming a 100 g sample, the mass of each element directly corresponds to its percentage. Thus:
  • Copper: 66.75%
  • Phosphorus: 10.84%
  • Oxygen: 22.41%
Understanding percentage composition helps in various applications, such as nutritional analysis and quality control in manufacturing.
Grasping the Concept of Molar Mass
Molar mass is the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a compound's formula according to the periodic table. Molar mass serves as a bridge between the macroscopic world we observe and the microscopic world of atoms and molecules.
For the determination of empirical formulas, as shown in the exercise above, knowing the molar mass of each component element is crucial. Here are the molar masses for the elements in our example:
  • Copper: 63.55 g/mol
  • Phosphorus: 30.97 g/mol
  • Oxygen: 16.00 g/mol
To determine the number of moles from a given mass, we divide the given mass by the molar mass of the element. The concept of molar mass is vital because it allows scientists to work with interconvertible mass and molar quantities, thus linking chemical formula information to real-world chemical compositions.

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Most popular questions from this chapter

Find the item in column 2 that best explains or completes the statement or question in column 1. Column 1 (1) 1 amu (2) 1008 amu (3) mass of the "average" atom of an element (4) number of carbon atoms in \(12.01 \mathrm{g}\) of carbon (5) \(6.022 \times 10^{23}\) molecules (6) total mass of all atoms in 1 mol of a compound (7) smallest whole-number ratio of atoms present in a molecule (8) formula showing actual number of atoms present in a molecule (9) product formed when any carbon-containing compound is burned in \(\mathrm{O}_{2}\) (10) have the same empirical formulas, but different molecular formulas Column 2 (a) \(6.022 \times 10^{23}\) (b) atomic mass (c) mass of 1000 hydrogen atoms (d) benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\), and acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\) (e) carbon dioxide (f) empirical formula (g) \(1.66 \times 10^{-24} \mathrm{g}\) (h) molecular formula (i) molar mass (j) 1 mol

Calculate the mass in grams of each of the following samples. a. 0.000471 mol of carbon monoxide b. \(1.75 \times 10^{-6}\) mol of gold(III) chloride c. 228 mol of iron(III) chloride d. 2.98 millimol of potassium phosphate (1 millimol = 1/1000 mol) e. \(2.71 \times 10^{-3}\) mol of lithium chloride f. 6.55 mol of ammonia

The molar mass of a substance can be obtained by ______ the atomic weights of the component atoms.

Calculate the molar mass for each of the following substances. a. ammonium sulfide, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{S}\) b. dichlorophenol, \(\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{OCl}_{2}\) c. barium hydride, \(\mathrm{BaH}_{2}\) d. potassium dihydrogen phosphate, \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) e. potassium hydrogen phosphate, \(\mathrm{K}_{2} \mathrm{HPO}_{4}\) f. potassium phosphate, \(\mathrm{K}_{3} \mathrm{PO}_{4}\)

Calculate the mass of carbon in grams, the percent carbon by mass, and the number of individual carbon atoms present in each of the following samples. a. 7.819 g of carbon suboxide, \(\mathrm{C}_{3} \mathrm{O}_{2}\) b. \(1.53 \times 10^{21}\) molecules of carbon monoxide c. 0.200 mol of phenol, \(C_{6} \mathrm{H}_{6} \mathrm{O}\)
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