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Chapter 9: Problem 127

Chrome yellow, a pigment used in paints, is \(64.11 \%\) by mass \(\mathrm{Pb}, 16.09 \%\) by mass \(\mathrm{Cr}\), and \(19.80 \%\) by mass O. What is the empirical formula of this compound?

Short Answer

Expert verified
The empirical formula of chrome yellow pigment, which consists of 64.11% lead (Pb), 16.09% chromium (Cr), and 19.80% oxygen (O) by mass, is \(PbCrO_4\).

Step by step solution

01

Convert percentages to mass

We can assume that there is 100 g of the pigment, as the given percentages will directly correspond to the mass in grams. Therefore, we have: - 64.11 g of lead (Pb) - 16.09 g of chromium (Cr) - 19.80 g of oxygen (O)
02

Convert mass to moles

To convert the mass of each element into moles, we should divide its mass by its molar mass. - The molar mass of Pb is \(207.2 g/mol\) - The molar mass of Cr is \(51.996 g/mol\) - The molar mass of O is \(16.00 g/mol\) Moles of Pb = \(\frac{64.11}{207.2} = 0.3091 \:mol\) Moles of Cr = \(\frac{16.09}{51.996} = 0.3091 \:mol\) Moles of O = \(\frac{19.80}{16.00} = 1.2375 \:mol\)
03

Find the smallest whole number ratio

Divide each moles value by the smallest moles value (0.3091). This gives: - Moles of Pb: \(\frac{0.3091}{0.3091} = 1\) - Moles of Cr: \(\frac{0.3091}{0.3091} = 1\) - Moles of O: \(\frac{1.2375}{0.3091} = 4\) Thus, the smallest whole number ratio is Pb:Cr:O = 1:1:4
04

Write the empirical formula

The empirical formula of this compound is: \(Pb_1Cr_1O_4 \), which can be simplified to: \(PbCrO_4\) So, the empirical formula of chrome yellow pigment is \(PbCrO_4\).

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