The question is slightly ambiguous because the term "empirical molecular weight" is not standard; however, we can infer it means "how to determine the empirical formula, which can lead to calculating a molar mass (or the mass of a mole) based on that empirical formula." Here is how you determine an empirical formula and how this is related to an empirical molar mass:
Determining the Empirical Formula
The empirical formula represents the simplest whole-number ratio of atoms in a compound. To find it, follow these steps based on the provided references:
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Convert Masses to Moles: First, if you're given the mass of each element in a compound, convert these masses to moles using the molar mass from the periodic table. You find the molar mass of each element from the periodic table.
- Example: If you have 12.01 grams of Carbon (C), you would divide this by the molar mass of Carbon (12.01 g/mol), to get 1 mole of Carbon.
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Calculate Mole Ratio: Once you have the moles of each element, divide each mole value by the smallest number of moles calculated. This step establishes the ratio of moles of each element.
- Example: Let's say you calculate 1 mole of Carbon and 2 moles of Hydrogen. Dividing both numbers by 1 (the smallest value), you get a 1:2 ratio of carbon to hydrogen.
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Round to Whole Numbers: Finally, round the resulting mole ratios to the nearest whole number. These whole numbers become the subscripts in the empirical formula, representing the ratio of elements in the compound.
- Example: Using the previous example of a 1:2 ratio, the empirical formula would be CH2.
Calculating Empirical Molar Mass
- Use the Empirical Formula Once you have the empirical formula (e.g. CH2) determine the molar mass for each element in it. For example, C = 12.01 g/mol, and H = 1.01 g/mol.
- Multiply by Subscript: Now multiply each of these atomic molar masses by their subscript in the empirical formula. For example, 12.01 g/mol 1 (from C1) + 1.01 2 (from H2).
- Add the Totals: Now sum all the products together. In this example 12.01 + 2.02 = 14.03 g/mol
- Answer: The result of all the steps is the empirical molar mass, in this case 14.03 g/mol.
Example:
Suppose we have a compound with 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
| Element | Mass Percentage | Molar Mass (g/mol) | Moles | Mole Ratio | Empirical Subscript |
|---|---|---|---|---|---|
| Carbon (C) | 40.0 g | 12.01 | 3.33 | 1 | 1 |
| Hydrogen (H) | 6.7 g | 1.01 | 6.63 | 2 | 2 |
| Oxygen (O) | 53.3 g | 16.00 | 3.33 | 1 | 1 |
Therefore, the empirical formula of this compound is CH2O.
- Empirical Molar Mass Calculation
- 12.01(C) + 1.01(H) * 2 + 16.00(O) = 30.03 g/mol.
- Thus the molar mass of the empirical formula CH2O is 30.03 g/mol.
This illustrates how you find the empirical formula, which then can be used to find the empirical molar mass.
