Molar Volume of a GAS at STP

• Gas density changes with pressure and temperature, and therefore it is important to use standards when referring to quantities of gas.
• We have a standard condition to compare volume of gasses called STP （STP stands for Standard Temperature and Pressure).
• Standard temperature is equal to 0 °C, which is 273.15 K.
• One mole of gas at STP occupies 22.4 liters

Exercise

Example 1
Calculate the number of mole of hydrogen gas that occupies a volume of 2.24L at S.T.P.
=> mole = 2.24/22.4 = 0.1

Example 2
Calculate the mass of oxygen gas(O₂) that expands to a volume of 33.6L at 0^oC and 1 atm pressure.
Step 1) Find the number of mole of oxygen gas.
=> mole = 33.6/22.4 = 1.5
Step 2) Find the mass of oxygen gas represented by 1.5 mole.
=> mass = mole X molar mass = 1.5 X 32 = 48 grams

Dilution of Solutions

When a solution is diluted, more solvent is added to it. 
    Since M = mole／V, and mole is the same for the original solution and the new diluted solution, it follows that M₁V₁ = M₂V₂ 
    where M₁=original concentration of solution 
              V₁=original volume of solution 
              M₂=new concentration of solution after dilution 
              V₂=new volume of solution after dilution


Examples

1.Calculate the new concentration (molarity) if enough water is added to 100mL of 0.25M sodium chloride to make up 1.5L.

M₂=(M₁V₁) ÷ V₂
M₁ = 0.25M
V₁ = 100mL = 100 ÷ 1000 = 0.100L (volume must be in litres)
V₂ = 1.5L
[NaCl(aq)]_new = M₂ = (0.25 x 0.100) ÷ 1.5 = 0.017M 
    (or 0.017 mol/L or 0.017 mol L^-1)

2.Calculate the volume to which 500mL of 0.02M coppper sulfate solution must be diluted to make a new concentration of 0.001M.

V₂=(M₁V₁) ÷ M₂
M₁ = 0.02M
V₁ = 500mL = 500 ÷ 1000 = 500 x 10^-3L = 0.500L (since there are 1000mL in 1L)
M₂ = 0.001M
V(CuSO₄)_new = V₂ = (0.02 x 0.500) ÷ 0.001 = 10.00L

Molarity: mole concentration

Molarity
-the number of moles of solute in one litre of a solution

In a homogenous mixture where one substance is dissolved in another
-the chemical that is being dissolved  is called solute
-the chemical that does the dissolving is called solvent


Formula



Exercise

How would you prepare 100. mL of 0.25 M KNO3 solution? Dissolve 2.53 g of KNO3 in enough water to make 100 ml of solution.

	A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution? 0.324 M
	How many moles of KBr are in 25.0 mL of a 1.23 M KBr solution? 0.0308 mol
	Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400. mL of this solution? 120 g

LAB 4C

LAB 4C FORMULA OF A HYDRATE






Determine the empirical formula of hydrate!
















Hydrate is compound that contains a definite number of water molecules incorporate into crystal structure.


An anhydrous salt is a salt that has been heated up until the water in it evaporates and it usually disenigrates into a powder, but is now anhydrous (Without water).

Procedure

• set up the pipestem triangle, iron ring, stand and bunsen burner
• heat up the crucible and cool it for 3 min
• place the hydrate in the crucible
• heating the crucible and cool it for 5 min
• reheating the crucible 
• note any change that occur

    *record the mass after each step*

Make a table to organize the information we get from experiment. By calculation, we can get the mass of the hydrate, anhydrous salt, and water given off. Since the teacher already gives us mass of one mole of anhydrous salt. It will be very easy to get the answer.

1. calculate the mole of anhydrous salt

mole(anhydrous salt) = mass of anhydrous salt/molar mass of anhydrous salt

2. calculate the mole of water

mole(water) = mass of water given off/ molar mass of water

3. divide each molar amount by the smallest molar amount we get just now

4. scall ratio to wholes number by multiplying

Now we are done!!!

Empirical formula of organic compound

 Empirical Formula of Organic Compounds

-The empirical formula of an organic compound (any substance that contains carbon) can be found by.

-Burning the compound (reaching it with O₂)

-Collecting and weighing the products

-From the mass of the products, the mole of each element in the organic compound can be calculated.

 

Example

    
What is the empirical formula of a compound that when a 500 gram sample is produces 15.0 grams of CO₂ and 8.18 grams of H₂O ?

-Step1: Find mole of CO₂ and H₂O

              mol CO₂ = 15.0g x 1mol/44.0g = 0.341mol   0.341mol/0.341mol = 1

              mol H₂O = 8018g x 1mol/18.0g = 0.454mol

              0.454 x 2 = 0.908899mol/0.341mole=2.66 2.66 x 3 = 8
Answer:C₃H₈

Empirical & Molecular formula

Calculating The Empirical and Molecular Formula

-Scientists use two kind of formula: molecular and empirical

 

Empirical Formula VS Molecular Formula

-Empirical formula of compound gives the lowest –term ratio of atom OR moles in the formula. All formula of Ionic compounds are empirical formulas.

-Molecular formula of a compound gives all atoms which make up a molecule. It is the formula of the ionic or covalent compound.

Examples

-Molecular Formula: C₆H₁₂O₆

-Empirical Formula:  CH₂O

- Molecular Formula:  Na₂C₂O₄

- Empirical Formula:  NaCO₂

Formula and Molecular Formula Converting between Empirical

- Molecular Formula = Empirical Formula x Whole number

- Molecular Formula Mass = Empirical Formula x Whole number

-Mass of one Mole = Empirical Formula Mass (in grams) x Whole number

Example

Find the molecular formula of a liquid with an empirical formula C₅H₁₂ and mass of one mole is 144.0g?

Step1: Calculate the empirical mass in grams of C₅H₁₂.

            5 x 12.0g + 12 x 1.0g =72.0g/mol

Step2: Since mass of 1 mole = empirical formula mass x Whole number

           144g / 72.0g=2

Answer: (C₅H₁₂) x 2 = C₁₀H₂₄

percent composition

 THE %BY MASS OF THE ELEMENTS IN A COMPOUND IS PERCENT COMPOSITION.

% composition=mass of element/mass of compound x 100%              

Emample  

1.Find the percentage composition of a compound that contains 1.94g of carbon, 0.48g of hydrogen and 2.58g of sulfur in a 5.00g sample of the compound.   

%C=1.94g/5.00g x 100%=38.8%

%  H=2.58g/5.00g  x 100%=51.6%

2.    An unknown compound with a mass of 34.5 grams contains 4.5g of hydrogen, 18g of oxygen and 12g of nitrogen.  What is percent composition of compound?

%H=4.5g/34.5g x 100%=13.0%

%O=18g/34.5g x 100%=52.1%

%N=12g/34.5g x 100%=134.8%