Molecular Formula - Video Tutorials & Practice Problems
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The molecular formula gives the actual number of atoms.
Molecular Formula
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concept
Molecular Formula
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recall that the molecular formula gives the actual number of different elements in a given compound, meaning it gives us the rial formula for the compound. The empirical formula, on the other hand, just gives us the relative number of those different elements in that same compound. Up to this point, we've learned to calculate the empirical formula when given either the masses or the percentages of those elements within that compound. But how exactly would we determine the molecular formula for those same compounds? Well, here we have a chart, and in his church we're talking about three compounds glucose octane and salicylic acid. We're going to say that they are empirical formulas are giving us ch 20 remember that there are sub scripts of one and one there we have C four, H nine and C 78603 In order to determine their molecular formulas, we can use end factors, and factors are just numbers that we can multiply the empirical formula by in order to determine the molecular formula. So, for example, glucose has an end factor of six. That means I will multiply each of the sub scripts of the empirical formula by six, and that will give me my molecular formula so that be six carbons, 12 hydrogen and six oxygen's. So the molecular formula glucose is C 68 12 Obtain. It has an n factor off to so multiplying those sub scripts gives me see eight h 18 and then finally, salicylic acid has a factor of one, which means that the molecular formula empirical formula are the same. So just remember, when you're given the end factor, just multiply the empirical formula by it and you'll determine what your molecular formula will be. But what happens when we don't have the end factor? How do we determine it in the next series of videos will go through the different steps required to find the different end factors for any given compound once you've determined its empirical formula.
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example
Molecular Formula Example 1
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the key. Defining your molecular formula is first determining your empirical formula. Now we're gonna say once the empirical formula is determined, the molecular formula can be obtained if the Moller Mass is also now realize here that the molar mass is connected to the molecular formula. If we take a look here in this example question, it says after workout session Lactic acid, which has a molar mass of 90.8 g per mole forms and muscle tissue and is responsible for muscle soreness. Elemental analysis shows that the compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen. Determine the molecular formula. All right. To determine the molecular formula, we follow step one in step one. It says Repeat the steps necessary to determine the empirical formula of the compound. Remember, What we do first is we're gonna convert all of these percentages immediately into grams because we assume that we're dealing with 100 g of our compound, so that's gonna be 40 g of C 6.7 g of H. 53.3 g of oxygen. Next, we're gonna convert each one of those grants into moles. So one more of each of these elements can be determined by finding their atomic masses. On the periodic table, One mole of carbon weighs 12.1 g. According to the periodic table, hydrogen is 1.8 g. Oxygen is 16 g. Here, all the grams cancel out and we'll have moles of each of the elements. So here we have 3.3306 moles of Sikh. And remember, when it comes to finding the moles at this point tow, Avoid rounding errors. Make sure you have at least four decimal places. Next we have one. We have 6.6468 moles of H and this is 3.3313 moles of O. Now, remember, at this point, we're gonna divide all the moles by the smallest small answer in order to determine the ratios to one another. The smallest mill answer that we got was three point 3306 So all of them will be divided by this number. So that's going to give me one carbon two hydrogen, one oxygen. So my molecular formula is C H 20 with the molecular formula for the empirical formula. Determined we're now going toe do Step two, Step two says. We calculate the empirical mass off the compound. The empirical mass comes from the empirical formula, which we just discovered, so that would be It's made up of one carbon to hydrogen, one oxygen multiply each one by their atomic masses from the periodic table. And then we're gonna add up those totals together. So we add those all together, we're gonna say the empirical mass of our compound is 30. g per mole. This takes us to step three. Step three says we divide the mower mask, which was given to us within the example. Problem off the molecular formula by the empirical mass to determine the end factor. So our n factor equals R. Mueller, Mass. Which was given to us within the question divided by the empirical mass which we can Onley determine once we find the empirical formula. So that would be 90.8 g per mole, divided by 30.26 g per mole, which gives us an end factor equal to three. Now we run a multiply the sub scripts off the empirical formula by the end factor to get the molecular formula. So here we're going to say that end factor times are empirical formula equals are molecular formula. Right. So then we're gonna plug in the three that we just found times the empirical formula, which you discovered earlier equals C three, 8603 This gives us the molecular formula for this question. So just remember the key to determine the molecular formula is the first Know what your empirical formula is once you have that determine your n factor and then n factor times empirical formula gives us our molecular form You
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Problem
Problem
What is the molecular formula for the following compound?
Empirical Formula: NPCl2 Molar Mass:347.64 g/mol
A
NPCl2
B
N3P3Cl6
C
N2P2Cl2
D
N4P4Cl8
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Problem
Problem
Cortisol (MW = 362.47 g/mol), a known steroid hormone, is found to contain 69.6% carbon, 8.34% hydrogen, and 22.1% oxygen by mass. What is its molecular formula?
A
C21H30O5
B
C21H26O5
C
C28H46O
D
C18H24O2
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Problem
Problem
Elemental analysis of a pure compound indicated that the compound had 72.2% C, 8.50% H and the remainder as O. If 0.250 moles of the compound weighs 41.55 g, what is the molecular formula of the compound?
