3.3 Mass percent and Emperical and molecular formulas Flashcards
Molecular formula of a compound
* basically just going to tell you exactly how many atoms are in each molecule
* EX: P4O10 <– in a single molecule you can see the exact atom
* EX: C3H8
Empirical formula: - a lot of times when you can’t reduce the formula this will be exactly the same as the molecular formula - true when your molecular formula has the most reduced integer ratio
* most reduced whole number formula
* P4O10 –> P2O5 (because they were both divisible by 2 we were able to reduce this) - has to be whole integers though
* C3H8 - same as molecular formula because it can’t be reduced anymore
Mass percent can be releated to both the emperical and molecular formulas
300gCaCO3
* so were going to explore this and find the percent of carbon in the sample and the percent of oxygen in the sample
Start by choosing a 1mol sample size because the mass percents are intensive - so you don’t even need to use your sample size # because its not important
1mol CaCO3
* 1 mol C per 1 mol CaCO3
* 1mol C = 12g
plug into easily derivable formula
12g C
—- x 100 = 12% carbon
100g
mass percents are an intensive property meaning it doesnt matter how big the sample is it will be the same
* this is because they’re a percent which is a ratio and with increased mass size both portions in the ratio grow proportionally keeping the percent stable
now find percent oxygen
* instead of using 300g CaCO3 I’m going to use 1mol
1mol CaCO3 = 3mol O = 48g O
48
– x 100 = 48% of the compound is oxygen
100 (total mass)
can also take mass percents and turn them into an emperical formula
You can also take mass percents and turn them into an emperical formula
A compound is 80% Carbon by mass, and 20% hydrogen by mass. What is its emperical formula?
* If the compound above has a molecular weight of 30amu what is its molecular formula?
So lets start by turning a mass percent into an emperical formula
* well since pass percent is intensive we can pick a number. Lets pick 100g
* 20% of 100g = 20g Hydrogen
* 80% of 100g = 80g Carbon
Our final formula is going to look like CxHy
* where x and Y are an integer ratio of # of atoms or moles of carbon to hydrogen
* Were going to look at it as a mole to mole ratio, not a gram to gram ratio
* so you can’t just say its C4H because thats a gram to gram ratio. Were looking for the mole to mole ratio
So lets start by converting out grams of carbon and hydrogen to moles
* shown below
This means that the formula is as far divided down as it can be
Meaning the weight is 50:50 - however the individual atoms dont weight the same amount so I need to figure out what percent of individual atoms are in the total mass to derieve a formula - using moles
So its importnat not to put SO because that would be correct if it was asking purely mass because half the mass is sulfur and the other half is oxygen
* hwoever, grams of different elements do not represent equal numbers of atoms.
* 1 mole of sulfur = 32.06 grams
* 1 mole of oxygen = 16.00grams
so, 50 grams of sulfur is not the same number of atoms as 50 grams of oxygen
Why we have to convert it to moles
You’re trying to find the simplest whole-number ratio of atoms, not grams
* you need to convert to moles to compare the number of particles (atoms)
* grams tells you how uch something weights, but moles tell you how many actual atoms or molecules you have
* moles unlock the ability to compare actual amounts of partials, which is why we always convert to moles when figuring out things like chemical formulas, reactions, and more.
Mass percents are intensive - meaning it doesnt matter what sample size you choose they will work out to the same %
So we assume a 1mol sample size of HNO3
Key is knowing its an intensive properpty and we can assume 1 mol
* remember the goal is to figure out what percent of the compounds mass is coming from Ti. So it doesnt matter how much of the compound we have, the percent of it that is Ti will be stable.
A few were really easy to rule out (when the other atom weighed tons)
However, I just needed to calculate the mass % for them (percent of the mass that was coming from iron)
FeO
I found that the emperical formula was NO2, however the answer was in the molecular formula which was N2O4 (basically the only formula that matched the NO2)
I started by assuming 100g because percents are intensive, meaning no matter how big that sample size was, its still going to be 30.4% N and 69.6% oxygen.
I then converted the grams of each atom into moles, so that I could compare the individual atoms instead of the grams (so that I could derieve the emperical formula) - atoms of nitrogen and oxygen weight different amounts - meaning if you compared their grams you’d be comparing different things
