Question

select all the correct answers. a compound has the percent composition 47.40% pd, 28.50% o, 21.40% c, and 2.69% h. based on this information, which molecular formulas could represent the compound? pd2c8h12o8 pdo4c2h9 pd(o2c2h3)3 pd(o2c2ch3)2 pdo2c2h3

Explanation:

Step1: Determine molar masses of elements

\(M_{Pd}=106.42\ g/mol\), \(M_O = 16.00\ g/mol\), \(M_C=12.01\ g/mol\), \(M_H = 1.01\ g/mol\)

Step2: Calculate molar mass and percent - composition of \(Pd_2C_8H_{12}O_8\)

Find molar mass, then percents.

Step3: Calculate molar mass and percent - composition of \(PdO_4C_2H_9\)

Find molar mass, then percents (not a match).

Step4: Calculate molar mass and percent - composition of \(Pd(O_2C_2H_3)_3\)

Find molar mass, then percents (not a match).

Step5: Calculate molar mass and percent - composition of \(Pd(O_2CCH_3)_2\)

Find molar mass, then percents (a match).

Step6: Calculate molar mass and percent - composition of \(PdO_2C_2H_3\)

Find molar mass, then percents (not a match).

Answer:

To solve this, we need to calculate the percent - composition of each given formula and compare with the given percentages.

1. First, find the molar masses of elements:

• Molar mass of Pd (palladium) \(M_{Pd}=106.42\ g/mol\)
• Molar mass of O (oxygen) \(M_O = 16.00\ g/mol\)
• Molar mass of C (carbon) \(M_C=12.01\ g/mol\)
• Molar mass of H (hydrogen) \(M_H = 1.01\ g/mol\)

1. For \(Pd_2C_8H_{12}O_8\):

• Molar mass of the compound \(M = 2\times106.42+8\times12.01 + 12\times1.01+8\times16.00\)
• \(M = 212.84+96.08+12.12 + 128.00=449.04\ g/mol\)
• Percent of Pd: \(\frac{2\times106.42}{449.04}\times100\%=\frac{212.84}{449.04}\times100\%\approx47.40\%\)
• Percent of O: \(\frac{8\times16.00}{449.04}\times100\%=\frac{128.00}{449.04}\times100\%\approx28.50\%\)
• Percent of C: \(\frac{8\times12.01}{449.04}\times100\%=\frac{96.08}{449.04}\times100\%\approx21.40\%\)
• Percent of H: \(\frac{12\times1.01}{449.04}\times100\%=\frac{12.12}{449.04}\times100\%\approx2.69\%\)

1. For \(PdO_4C_2H_9\):

• Molar mass of the compound \(M = 106.42+4\times16.00+2\times12.01+9\times1.01\)
• \(M = 106.42 + 64.00+24.02+9.09 = 203.53\ g/mol\)
• Percent of Pd: \(\frac{106.42}{203.53}\times100\%\approx52.38\%\) (not correct)

1. For \(Pd(O_2C_2H_3)_3\):

• First, the molar - mass of \(O_2C_2H_3\) is \(2\times16.00+2\times12.01 + 3\times1.01=32.00+24.02+3.03 = 59.05\ g/mol\)
• Molar mass of \(Pd(O_2C_2H_3)_3\) is \(106.42+3\times59.05=106.42 + 177.15=283.57\ g/mol\)
• Percent of Pd: \(\frac{106.42}{283.57}\times100\%\approx37.52\%\) (not correct)

1. For \(Pd(O_2CCH_3)_2\):

• Molar - mass of \(O_2CCH_3\) is \(2\times16.00+2\times12.01+3\times1.01=32.00 + 24.02+3.03=59.05\ g/mol\)
• Molar mass of \(Pd(O_2CCH_3)_2\) is \(106.42+2\times59.05=106.42+118.10 = 224.52\ g/mol\)
• Percent of Pd: \(\frac{106.42}{224.52}\times100\%\approx47.40\%\)
• Percent of O: \(\frac{2\times2\times16.00}{224.52}\times100\%=\frac{64.00}{224.52}\times100\%\approx28.50\%\)
• Percent of C: \(\frac{2\times2\times12.01}{224.52}\times100\%=\frac{48.04}{224.52}\times100\%\approx21.40\%\)
• Percent of H: \(\frac{2\times3\times1.01}{224.52}\times100\%=\frac{6.06}{224.52}\times100\%\approx2.69\%\)

1. For \(PdO_2C_2H_3\):

• Molar mass of the compound \(M = 106.42+2\times16.00+2\times12.01+3\times1.01\)
• \(M = 106.42+32.00+24.02+3.03 = 165.47\ g/mol\)
• Percent of Pd: \(\frac{106.42}{165.47}\times100\%\approx64.31\%\) (not correct)

The correct molecular formulas are \(Pd_2C_8H_{12}O_8\) and \(Pd(O_2CCH_3)_2\).