Molar Masses A = 17.813- mole D = 41.486- mole E = 33.011- mole Y = 19.280 X = 255.691 mole Z = 24.650 mole mole A compound was analyzed and found to contain 17.60 g D, 56.03 g E, and 16.36 g Y. The molecular weight of the compound is 424.177 Using the molar masses provided above please determine the empirical and molecular formula for the compound. mole Don't worry about subscripts, C2H60 would be entered as C2H60 Empirical Formula: type your answer. Molecular Formula: type your answer.

Chemistry
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Chapter1: Chemical Foundations
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**Molar Masses**

\[ A = 17.813 \, \frac{g}{mole} \]

\[ D = 41.486 \, \frac{g}{mole} \]

\[ E = 33.011 \, \frac{g}{mole} \]

\[ Y = 19.280 \, \frac{g}{mole} \]

\[ X = 255.691 \, \frac{g}{mole} \]

\[ Z = 24.650 \, \frac{g}{mole} \]

A compound was analyzed and found to contain 17.60 g of \( D \), 56.03 g of \( E \), and 16.36 g of \( Y \). The molecular weight of the compound is 424.177 \, \(\frac{g}{mole}\).

Using the molar masses provided above, please determine the empirical and molecular formula for the compound.

Don't worry about subscripts, \(\text{C}_2\text{H}_6\text{O}\) would be entered as \(\text{C2H6O}\).

- Empirical Formula: [Type your answer...]

- Molecular Formula: [Type your answer...]
Transcribed Image Text:**Molar Masses** \[ A = 17.813 \, \frac{g}{mole} \] \[ D = 41.486 \, \frac{g}{mole} \] \[ E = 33.011 \, \frac{g}{mole} \] \[ Y = 19.280 \, \frac{g}{mole} \] \[ X = 255.691 \, \frac{g}{mole} \] \[ Z = 24.650 \, \frac{g}{mole} \] A compound was analyzed and found to contain 17.60 g of \( D \), 56.03 g of \( E \), and 16.36 g of \( Y \). The molecular weight of the compound is 424.177 \, \(\frac{g}{mole}\). Using the molar masses provided above, please determine the empirical and molecular formula for the compound. Don't worry about subscripts, \(\text{C}_2\text{H}_6\text{O}\) would be entered as \(\text{C2H6O}\). - Empirical Formula: [Type your answer...] - Molecular Formula: [Type your answer...]
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