BTC

Percent CompositionEmpirical FormulasandStoichiometry: 

Percent Composition Empirical Formulas and Stoichiometry By: Brent Mauldin Taylor Wiggins and Caitlyn Imm

Percent Composition: 

Percent Composition is the percentage by mass of each element in a compound Take for example H2O. H2 makes up 11.1% of water and O makes up the other 88.9% Here is how we know...

Slide3: 

The original formula for water is H2O, right? Now look up the molar mass for each element on the periodic table. The molar mass of H is 1 and the molar mass of O is 16. Now since we have 2 Hs we need to multiply the molar mass by 2. Now we add the total mass. 2+16=18 Now we divide 2 by 18 to get a decimal we can convert to a percentage.

Slide4: 

2/18 is about .111. Now we multiply by 100 to get the percent. So water is 11.1% H. Now to find O, we only need to subtract H’s percentage from 100 So water is also 88.9% O. (Hint: to check yourself on real problems, just add all of the percentages and they should equal about 100)

Slide5: 

Here is the whole problem worked out: Step 1- Find molar mass (look at the PTE) H- 1x2=2 O- 16x1=16 Step 2- Find total molar mass 2+16=18 Step 3- Find the percentage 2/18~.111 16/18~.889 .111x100=11.1% .889x100=88.9% Step 4- The answer H-11.1% O-88.9%

Slide6: 

Now for another example… What is the percent composition of Cu2S? Step 1- Find the molar mass of each element (do NOT forget to look at the periodic table) Cu- 64x2=128 S- 32x1=64

Slide7: 

Step 2- Find the total molar mass (add everything up) 128+64=192 Step 3- Find the percentage 128/192~.667 64/192~.333 .667x100=66.7% .333x100=33.3% Step 4- The answer Cu- 66.7% S- 33.3%

Empirical Formula: 

Empirical Formula A chemical formula that shows the composition of a compound in terms of the relative numbers in kinds of atoms in the simplest ratio is the empirical formula. For example, the empirical formula of C4H8 is CH2 This is how we know…

Slide9: 

The ratio of the number of C atoms to H atoms in C4H8 is 4:8. When we simplify the ratio, we get 1:2, so the empirical formula is CH2 Easy right? Well, don’t worry it gets more complicated.

Slide10: 

Find the empirical formula of C6H8O6 All we do is set up another ratio, this time between carbon, hydrogen, and oxygen. The ratio is 6:8:6. When we simplify it, we get 3:4:3. Thus, the empirical formula is C3H4O3.

Slide11: 

Now let’s do a problem where we use percent composition and empirical formula. A 10.150 g sample of a compound that only contains phosphorus and oxygen contains 4.433 g phosphorus. What is the empirical formula?

Slide12: 

We will first find the amount of oxygen by subtracting the amount of phosphorus from the total mass: 10.150-4.433=5.717g O Now we have to convert grams to moles because you can’t compare grams to grams, only moles to moles. 4.433/30.97=.1431 mol P 5.717/16.00=.3573 mol O

Slide13: 

Next, we simplify the mole ratio 1 mol P : 2.5 mol O Next, multiply by 2 to get… 2 mol P : 5 mol O Which means that the empirical formula is… P2O5

Stoichiometry: 

Stoichiometry a branch of chemistry that deals with the application of the laws of definite proportions and of the conservation of mass and energy to chemical activity Sounds confusing, right? Don’t worry, we will do some example problems to get you started.

The End: 

The End The stoichiometry problems are on the video. After you watch the video, please take the quiz and do the worksheet on the wikispace.