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Chapter 8: Problem 64

If cobalt metal is mixed with excess sulfur and heated strongly, a sulfide is produced that contains \(55.06 \%\) cobalt by mass. Calculate the empirical formula of the sulfide.

Short Answer

Expert verified
The empirical formula of the cobalt sulfide compound is Co鈧係鈧�.

Step by step solution

01

Find the mass percentage of sulfur

Since the cobalt sulfide compound consists of cobalt and sulfur only, the mass percentage of sulfur can be found by subtracting the mass percentage of cobalt from 100%: Mass percentage of sulfur = 100% - mass percentage of cobalt = 100% - 55.06% = 44.94% So, we now have the mass percentages of cobalt (55.06%) and sulfur (44.94%) in the compound.
02

Convert mass percentages to grams

To convert the mass percentages into grams, we'll assume we have 100 grams of the compound. In this case, the mass of cobalt and sulfur would be equal to their mass percentages: Mass of cobalt = 55.06g Mass of sulfur = 44.94g
03

Convert grams to moles

To find the mole ratio, we need to convert the grams of cobalt and sulfur to moles. To do this, we'll use the molar mass of each element: Moles of cobalt = mass of cobalt / molar mass of cobalt Moles of cobalt = 55.06g / 58.93g/mol = 0.934 mol (rounded to three decimal places) Moles of sulfur = mass of sulfur / molar mass of sulfur Moles of sulfur = 44.94g / 32.07g/mol = 1.402 mol (rounded to three decimal places)
04

Determine the mole ratio

To find the empirical formula, we'll determine the mole ratio between cobalt and sulfur. To do this, we'll divide both mole values by the smallest mole value: Mole ratio of cobalt = 0.934 mol / 0.934 mol = 1 Mole ratio of sulfur = 1.402 mol / 0.934 mol = 1.5 (rounded to one decimal place) Since empirical formulas should be expressed with whole numbers, we can multiply both ratios by a suitable number to convert the ratios into whole numbers. In this case, we'll multiply both ratios by 2. Mole ratio of cobalt = 1 脳 2 = 2 Mole ratio of sulfur = 1.5 脳 2 = 3
05

Write the empirical formula

Now that we have the whole number mole ratios for cobalt and sulfur, we can write the empirical formula: Empirical formula: Co鈧係鈧� So, the empirical formula of the cobalt sulfide compound is Co鈧係鈧�.

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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

Give the empirical formula that corresponds to each of the following molecular formulas. a. sodium peroxide, \(\mathrm{Na}_{2} \mathrm{O}_{2}\) b. terephthalic acid, \(\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}\) c. phenobarbital, \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{3}\) d. 1,4 -dichloro-2-butene, \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{Cl}_{2}\)

Calculate the percent by mass of each element in the following compounds. a. calcium phosphate b. cadmium sulfate c. iron(III) sulfate d. manganese(II) chloride e. ammonium carbonate f. sodium hydrogen carbonate g. carbon dioxide h. silver(I) nitrate

If an average sodium atom weighs 22.99 amu, how many sodium atoms are contained in \(1.98 \times 10^{13}\) amu of sodium? What will \(3.01 \times 10^{23}\) sodium atoms weigh?

Calculate the mass in grams of each of the following samples. a. 2.41 millimol of potassium nitrate (1 millimol \(=\) \(1 / 1000 \mathrm{mol})\) b. 8.91 mol of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) c. 0.0141 mol of calcium oxide d. \(1.91 \mathrm{mol}\) of gold(III) bromide e. 0.0000117 mol of water f. 2.68 mol of silver nitrate
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