In this video, we're going to take a look at the difference between our empirical formula and molecular formula. First of all, when it comes to the empirical formula of a compound, we're going to say it's related to the mass percentage of its constituent elements using the mole concept. Remember, the mole concept has to do with us going between grams, moles, and units such as molecules, ions, or atoms. We're going to say here that when it comes to our compound, the empirical formula gives us the relative number of atoms and represents the most simplified form. So by convention, we're going to say any formula must contain whole numbers of each atom in what's called the whole number ratio. Now let's put this to practice and really understand what's the difference between molecular formula and empirical formula. Here, we have the molecular formulas of different compounds. This is how many actual elements of each type that are present within that compound. The empirical formula can be thought of as its reduced or simplified form. So if we take a look here, we have C3H6O3. Realize that here 3, 6, and 3 are all divisible by 3. So if I divide this by 3, what I'll have left is CH2O. That CH2O represents the reduced simplified form or the empirical formula. If we look at the next one, we have C10H14N2. All these numbers here are divisible by 2. So when we divide everything by 2, we're going to get C5H7N. That represents the empirical formula. And then finally, we have C12H22O11. They don't share any number in common where we can reduce it to a simplified form, so we'd actually have the same empirical formula as molecular formula, and we'll see from time to time that this is true. So just remember that your molecular formula is the actual number of each of the elements within a compound, and the empirical formula is the reduced form of that compound. Keep this in mind when comparing molecular formula versus empirical formula.
Empirical Formula - Online Tutor, Practice Problems & Exam Prep
The empirical formula gives the relative number of atoms.
Empirical Formula
Empirical Formula
Video transcript
Empirical Formula Example 1
Video transcript
We're now going to calculate the empirical formula for a given compound. The empirical formula can be calculated from the masses or percentages of elements within a compound. In this example question, it says, "Determine the empirical formula of a compound that is 68.40 percent chromium and 31.60% oxygen."
Step 1, you're going to write down the symbols for each element in the question. We're dealing with chromium and its symbol as Cr and oxygen with its symbol as just O.
Step 2, we're going to write down the masses in grams of each element given. Notice here in this question that they're not giving us masses. What they're giving us are percentages. We're going to convert all the percentages into grams by assuming there are 100 grams of the compound. If you add the percentages up, 68.40 percent plus 31.60 percent, you'll see that you get 100% of your total compound. This makes sense because if our compound is made up of chromium and only oxygen, together they should add up to the complete compound. So we're going to convert 68.40% to 68.40 grams of Cr, and 31.60 grams of just O.
Step 3, convert all the masses into moles. This requires some mole concept theory on your part. For chromium, 68.40 grams of Cr requires conversion, using that 1 mole of Cr weighs 51.996 grams, giving us 1.3155 moles of Cr. For oxygen, 31.60 grams of oxygen requires conversion using that 1 mole of oxygen weighs 16.00 grams, resulting in 1.9750 moles of oxygen. To avoid rounding errors, ensure the values have at least 4 decimal places.
Step 4, you're going to divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. Here we have 1.3155 moles of Cr and 1.9750 moles of O. Dividing both by 1.3155, we get 1 chromium and 1.5 oxygens.
Step 5, after dividing by the smallest number, if you encounter a fraction such as 0.5, which can't simply be rounded, we multiply by a factor to create whole numbers. In this case, multiplying by 2, we get 2 chromiums and 3 oxygens, rendering the empirical formula as Cr2O3.
These steps must be employed to determine the empirical formula when given information on percentages or mass of different elements within a compound.
A compound that contains only carbon, hydrogen, and oxygen is composed of 48.64% C and 43.2% O by mass. What is the empirical formula of this compound?
Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula of the compound?
A compound composed of carbon, hydrogen, and chlorine contains 4.19 x 1023 hydrogen atoms. If 9.00 g of the compound also contains 55.0% chlorine by mass, what is the empirical formula?