To compare the amount of solute (chemical being dissolved) in an amount of solution (chemical doing the dissolving), we will need to use the formula: Moles of solute (mol)/Volume of solution (L)
This formula will give us the molarity or molar concentration of the solution. (M)
Using the formula M= mol/L, we can find:
a) the amount of moles or grams of the solute
b) the molarity of the solution
c) the amount of solution in Litres
Below are example questions that might help you practice:
Find the molarity of a solution with 0.5 moles of sodium chloride dissolved in 0.75 L of water
M = mol/volume
=0.5/0.75
= 2/3 M of sodium chloride
Find the amount of solution present if the molarity is 1.2M and there is 3.6 moles of lithium sulphate is present in the solution.
M = mol/volume
volume = mol/M
volume = 3.6/1.2
volume = 3 Litres
Today, we had a lab on determining the percentage and moles of water in an unknown hydrate and writing the empirical formula of the hydrate. Basicly, all we did was to heat the unknown hydrate and weight the remaining hydrate to see how much water was given off by the hydrate. Also, we added a few drops of water to the remainings to see how the colour of hydrate have changed.
Here's the brief procedure of the lab. 1. Set up the equipments (crusible, pipestem triangle, iron ring, & bunsen burner) as shown on the right. Heat the crusible to make sure that it is dry. After cooling the crusible, determine & record the mass of the empty crusible.
2. Place the hydrate into the crusible, determine & record the mass.
3. Heat the crusible with hydrate in it for about 5 min. Turn off the burner,let it cool, and determine & record the mass of it.
4. Reheat the crusible with hydrate to make sure that all the water is driven off.
5. Add a few drops of water to the comtents of the crusible and note any changes occured.
A: Determining the percentages and moles of the water.
To determine the percentages of water, you need to know the mass of water given off by the hydrate. Simply, when you take the mass of the hydrate before heating and subtract the mass of the hydrate after heating (which is anhydrous salt ), that's the mass of water given off by the hydrate. Since you know the mass of hydrate and water, you can calculate the percent composition of the water. ( For calculating percent composition of a compound, see the previous post! )
% water = mass of water / mass of hydrate ｘ 100
Since you know the mass of water, it is easy to determine the moles of water present in hydrate! ( See the previous post for steps to calculate the moles of substance in a compound!)
number of moles of water = mass of water x 1 mole water / 18.0g water
B: Determining the empirical formula of the hydrate.
In order to determine the empirical formula, you need to know the moles of anhydrous salt present in hydrate. Take the same steps as part A to deter mine the number of moles of anhydrous salt.
Once you got the moles of water and anhydrous salt, you can determine the empirical formula drom finding the ratio of these two substances in the hydrate by making the number of moles into whole numbers. ( see the previous post!)
★These pictures are showing the possible colour change that can occur when heating a hydrate.
This video will explain more about hydrates and their formulas
An organic compound is any substance that contains carbon. ( It is important to identify this on a question)
The moles of each element in the organic compound can be calculated from the mass of the products.
This is the method we use to find the empirical formula of an organic compound, and here is the equation:
CxHy + zO2 + xCO2 + y/2 H2O
This is the balanced chemical equation for the burning of CxHy.
The moles of C and H in CxHy = moles of C in CO2 and the moles of H in H2O.
Now let's apply this into an example problem.
Ex/ When 3.79 grams of an organic compound is burned, 6.61 grams of CO2 and 3.59 grams of H2O is produced. What is the empirical formula of this compound?
Step 1: Find moles of CO2 and H2O
6.61 grams CO2 x 1 mol CO2 / 44 grams CO2 = 0.150 mol CO2
3.59 grams H2O x 1 mole H2O / 18 grams H2O = 0.199 mol H2O
Step 2: Find moles of C and H
0.150 x 1mol C/1 mol CO2 = 0.150 mol C
There is one C in CO2, so we multiply by 1 getting the same result but changing the unit.
0.199 x 2 mol H/1 mol H2O = 0.399 mol H
As for H, there are 2H in H2O so we simply multiply by 2.
Step 3: Check masses
If the mass of C and H does not equal the mass of the compound, then there must be a component of Oxygen present in the compound.
Therefore, mass of Oxygen = mass of compound - mass of C and H
0.150 mol C x 12 grams C / 1 mol C = 1.80 grams C
0.399 mol H x 1 grams H / 1 mol H = 0.399 grams H
1.80 grams + 0.399 grams = 2.199 grams
Now we need to find out the mass of Oxygen.
3.79 grams - 2.199 grams = 1.591 grams O
1.591 grams O x 1 mol O / 16 grams O = 0.099 mol O
Step 4: Find ratio + change ratio into whole number
Just like what we have learned in the previous lesson, divide each molar amount by the smallest molar amount.
C: 0.150 mol / 0.099 mol = 1.5
H: 0.399 mol / 0.099 mol = 4
O: 0.099 mol / 0.099 mol = 1
Now we need to make these into whole numbers.
C: 1.5 x 2 = 3
H: 4 x 2 = 8
O: 1 x 2 = 2
Final answer: C3H8O2
The above is a pretty good video to review the last lesson as well as what we learned in this lesson.
