Molarity

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

For more practice problems~click  here to go to a web with a molarity worksheet
Answers are also included!

Lab 4-C Formula of a Hydrate

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

Empirical Formula and Organic Compounds

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.

Percent Composition of Molecules

When you find a percent composition of molecules, you calculate the molar mass of an element, which is shown under each element on the periodic table, then calculate each element's percentage of that total.

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

Example:
What is the % composition of H2O ?

First, calculate the total molar mass of H2O

Molar mass of 2 hydrogen is 2.0 g/mole . Molar mass of oxygen is 16.0 g/mole.


So, the total molar mass of H2O is 2.0 + 16.0 = 18.0 g/mole

% H = 2.0 / 18.0 x 100% = 11.1 %
% O = 16.0 / 18.0 x 100% = 88.9 %

The results together should be added up to 100%.

Empirical and Molecular Formulas

Empirical Formula: The ratio of the number of elements in a compound expressed in lowest form
Ionic Compounds are usually an example of empirical formulas

Example: NaCl

               MgO

Molecular Formula: The actual amount of the number of elements in a specific compound
Some covalent compound's molecular formula is not the same as their empirical formula
Example: C₄H₁₀  (The empirical form would be C₂H₅)               H₂O₂   (The empirical form would be HO)To convert from and Empirical to Molecular or Molecular to Empirical, we need to find the whole number to multiply it by.

Converting Between Empirical and Molecular Formulas:

The three formulas are:

Molecular Formula = Empirical × Whole Number (N)

Mass of 1 mole = Empirical mass in grams × Whole Number

Molecular Formula Mass = Empirical × Whole Number

If the empirical formula of a compound is C₂H₅and the molar mass is 58g/mol, what is the molecular formula for this compound?
First, we find the mass the C2H5 (Empirical Formula)
C = 12.0x2 = 24.0g
H = 1.0x 5 = 5.0g
24.0+5.0 = 29.0g


Then, divide the molar mass by the mass of the empirical formula to calculate the whole number (N) used to separate between the emiprical and the molecular formulas.


58g/mol ÷29.0 g = 2 

Find the empirical formula of a compound that contains 7.2grams of Carbon, 1.2grams of Hydrogen, and 9.6grams of Oxygen.

Step 1: Convert the grams of elements into moles

C:  7.2g C×1 mol C/12.0g C = 0.6 mol

H:  1.2g H×1 mol H/1.0g H = 1.2 mol

O:  9.6g O×1 mol O/16.0g O = 0.6 mol

Step 2: Divide the moles by the smallest amount of mole to get the empirical formula

C:  0.6mol C/0.6mol C = 1

H:  1.2mol H/0.6mol C = 2

O:  0.6mol O/0.6 molC = 1

Therefore, the empirical formula for this compound would be: CH₂O

For further examples, here is a video to show you the some conversions between empirical and molecular formulas.

Two-step mole conversions

When converting perticles into grams or grams into particles, use two-step mole conversion.

Converting PARTICLES into GRAMS
Ex. What is the mass of 7.32ｘ10¹⁵ Tin(Sn)　atoms?
      1.Convert ATOMS into MOLES

          7.32ｘ10¹⁵Sn atoms ｘ 1mole / 6.022ｘ10²³ atoms
                        = 1.21ｘ10⁻⁸Sn moles
                          (always remember SIG FIGS!)

      2. Then, convert MOLES into GRAMS

         According to the periodic table, 1 mole of Sn = 118.7g
         1.21ｘ10⁻⁸Sn moles ｘ 118.7g / 1mole
                         = 143.627 g of Sn

     So, the mass of 7.32ｘ10¹⁵ Tin atoms is 144 grams.


Converting GRAMS into PARTICLES
Ex. How many atoms of Hydrogen are present in 23.0g of NH₃?

1. Convert GRAMS into MOLES

          Since 1mole of NH₃=  17.0g,
          23.0g of NH₃ ｘ 1mole / 17.0g
                       = 1.35 moles of NH₃

     2. Then, convert MOLES into PARTICLES

　　　　 1.35 moles of NH₃ｘ 6.022ｘ10²³ molecules/ 1mole
                       = 8.13ｘ10²³ molecules of NH₃
           8.13ｘ10²³ molecules of NH₃ｘ  3H atoms / 1 molecule NH₃
                       = 2.44ｘ10²⁴H atoms

     So, 2.44ｘ10²⁴atoms of Hydrogen are present in 23.0g of NH₃.

Here are some mole jokes that you might enjoy~!

Why are moles bad at counting?

Because they only know one number.
What's the mole's favorite brand of soda?
Coca-Mola.

Why are moles always on the phone?
Because they love moleble devices.
Why do moles love Tyra Banks?
Because she's on America's Next Top MoledelWho is the the mole's favorite actor?
Mole Gibson

What does Avogadro put in his hot chocolate?
Marsh-mole-ows!

Mole Conversions

The above diagrams shows The conversion steps between Grams, Moles and Particles.

Now we are going to try to apply these:

(Remember sig figs!!)

ex/ How many moles of CCl4 are present in 8.36 x 10^35 molecules of CCl4?

Particle/Atom/Formula Unit ---> Moles

8.36 x 10^35 x 1/6.022 x 10^23 = 1.39 x 10^12 moles

ex/ How many copper atoms are present in 3.1 x 10^-3 mole of Cu2SO4

Moles ---> Particle/Atom/Formula Unit

Here it gets a little tricky, you are asked to find the number of copper atom, but it is Cu2, therefore you're gonna have to times 2 to get your final result.

3.1 x 10^-3 x 6.022 x 10^23 x 2 = 3.7 x 10^21 copper atoms

ex/ What is the mass in grams of 6.663 moles of neon gas?

Moles ---> Grams

Molecular mass of Ne2 = 2 Ne x 20.2 = 40.4 u

Molar mass of Ne2 = 40.4 g/mole

6.663 x 40.4 g/mole = 269 g Ne2

ex/ How many moles are there in 92.0 grams of Lead?

Grams ---> Moles

Atomic mass of Pb = 207.2 u

Molar mass of Pb = 207.2 g/mole

92.0 x 1/207.2 = 0444 mol Pb

Moles

Mole is a unit like milometers, seconds, or litres.

A long time ago, scientists studied different gases and determined their properties.

They found the masses of hydrogen, oxygen, and carbon dioxide.

By finding the mass, they also discovered that if the gases had the same volume, they have the same ratios

Oxygen : Hydrogen = 16:1

Carbon Dioxide : Hydrogen = 22:1

Carbon Dioxide : Oxygen = 11:8

Different Types of Masses:

Relative Mass: a comparison (mathematically) between the mass of two objects

Hydrogen and Oxygen were all previously used as a standard for comparison but now, we use CARBON (mass= 12amu)as the standard for comparison.

Avogadro is a scientist that recognized the pattern and created:

 AVOGADRO'S HYPOTHESIS

Equal volumes of different gases at the same temperature and pressure have he same number of particles

If the number of particles are the same, that means the mass ratio is due to the mass of the particles.

This principle was then used to create the periodic table.

Atomic Mass: mass of 1 atom of an element in atomic mass units

Example: Sodium has an atomic mass of 23.0 amu

Formula Mass: Mass of all the atoms in an ionic compound

Example: NaCl (salt)

Na = 23.0 amu

Cl = 35.5 amu

23.0 + 35.5 = 58.5 amu

Molecular Mass: Mass of all the atoms in a covalent compound

Example: CO

Carbon = 12.0 amu

                                                        Oxygen = 16.0 amu

                                                       12.0 + 16.0 = 28.0 amu

Molar Mass: Mass of one mole of each element

Example: 1 mole of oxygen = 16.0g/mol

1 mole of fluorine = 19.0 g/mol

Avogadro's number : 6.022x10^23

This is the number of particles in 1 mole of a substance.

The mole is just a unit like kilometers and meters and when it is used to express the molar mass of an atom, the unit for mole is grams/mol.

The mole is important because it allows chemists to use their mass to calculate how much atoms and molecules there are instead of actually counting them.

To give you an idea of how big one mole is, here is a video.

                   

Lab 2E

Objectives: To calculate the thickness of a sheet of aluminum foil and express the answer of proper scientific notation and significant figures.

Equipment

3 rectangular pieces of aluminum foil ( minimum 15 cm x 15 cm )

metric ruler

centigram balance

Procedure

Put the rectangular pieces of aluminum foil on each side and write number 1-3 on each respectively. Measure the length and width of each piece with a metric ruler. Record measurements ( using sig figs ) on table 1. Find the mass of each piece with a centigram balance and record it. Finally, compare and discuss results with groupmates.

 Sheet       Length ( cm )      Width ( cm )       Mass ( g )
   1               15.5                     14.6               0.88
   2               16.7                     15.1               1.05
   3               14.5                     14.9               1.02


The thickness of each foil   2.7 = density      ( D = M/V )
1 ) 2.7 = 0.88 / V                            2.38 = 15.5 x 14.6 x h
      V = 2.38 cm cubed                         h = 1.05 x 10^-2
2 ) 2.7 = 1.05 / V                            0.39 = 16.7 x 15.1 x h
      V = 0.39 cm cubed                         h = 1.55 x 10^-3
3 ) 2.7 = 1.02 / V                            0.38 = 14.5 x 14.9 x h
      V = 0.38                                         h = 1.8 x 10^-3

Average thickness : ( 1.05 x 10^-2 + 1.5 x 10^-3 + 1.8 x 10^-3) / 3
                            = 4.6 x 10^-3

Accepted value = 1.55 x 10^-3
                 * sheet 2 has the most accurate result.

Here is a QUIZ to help you further understand mass volume and density and how to apply them to questions.

Excel and Graphing

Today, we learned about how to graph data using excel.

How to create a graph:
1. Put in the data into the cell boxes in excel

2. highlight all the cells with the data and go to plot scatter graph.

This will give you a graph that looks like this:

3. Make sure to add a title, and lable the axis with the correct units.

4.  If the dots seem to make a straight line, it is possible to draw a "best fit" line through it and show the equation for the graph.

If these four simple steps are followed, it will be really easy to make graphs using excel when there is a set of data that needs to be graphed.



Density

Density is a physical property of matter that describes the amount of mass to a unit of volume.

We use the formula below to calculate the density.



 In this formula, D equals M divided by V where D stands for Density, M stands for mass, and V stands for Volume.

Some of the possible units for density are; g/mL, g/L, g/cm³, kg/L

 If an object is less dense than the liquid, it SINKS

 

If an object is more dense than the liquid, it FLOATS

Here's the video showing what happens if you put an egg into two liquids with different density(water&salt water)

 

 

Example Question:

A substance has a mass of 44.1g and a volume of 12.6cm³. What is it's density?

 

Use the formula to calculate the density.

　　　D=M/V

  　　　 =44.1g/12.6cm³

   　　　=3.5g/cm³

So, its density is 3.5g/cm³.

 

 

 

Here's something you can do using different densities the substances have.

 

 

 

 

 

 

Measurement and Uncertainty

No Measurement is exact,

measurement is a best estimate which has some degree of uncertainty.

We can only be certain of an amount when we count.

 

Here are some of the methods to determine the degree of uncertainty:

Absolute Uncertainty

Expressed in the units of measurement, not as a ratio.

Method 1: Make at least 3 measurements and calculate average.

                  The absolute uncertainty is the largest different between the average and lowest or

                  highest reasonable measurement. Discard the unreasonable data.

e.g. 5.9, 5.8, 5.9, 6.0, 6.5 <- the 6.5 here would be an unreasonable data that we don't need

now find the average: 5.9+5.8+5.9+6.0 = 23.6 -> 23.6 / 4 = 5.9

final answer: 5.9 ± 0.1 <- absolute uncertainty

Method 2: Determine uncertainty of instrument

                   Always measure to the best precision,

                   estimate to a fraction 0.1 of the smallest segment of the instrument scale

e.g. Smallest division: 1 cm, measure to 0.1 cm

Relative Uncertainty = Absolute Uncertainty ÷ Estimated Measurement x 100

a ± b  ->  b ÷ a x 100 = ___% relative uncertainty

Or use sig figs,

 number of sig figs indicates relative uncertainty

Here is a video to help understand this lesson better:

Significant Figures and Rounding

Accuracy -  how close the measurement is compared to the actual value
Precision - to what value does measuring the same object under the same conditions show the same result

An instrument can be very precise but not accurate
Example- a ruler could have precision down to the milimeter but if the ruler does not have to necessarily be accurate with its measurements.

Significant Figures or Meaningful Digits

• The more significant figures a measurement has, the more precise the measurement is
• The last significant digit in the measurement is uncertain
• All of the other significant digits are certain
• Significant digit contains all the certain digits and only the FIRST uncertain digit
• Any zero between a decimal and a significant digit do not count if the number is smaller than 1
• Zeros after significant digits count if there is a decimal
• Zeros between two significant digits also count

This example shows when zeros do not count as significant digits

•  0.00023
• Only 2 and 3 count as significant digits
• The four zeros in front DO NOT count  

These zeros count as significant digits

• 3000 = 1 significant digit
• 1.01 = 3 significant digits
• 1.1000 = 5 significant digits

To help you understand scientific notation better, here is a QUIZ on it:

Exact Numbers/ Rounding

Some numbers are exact and so are not to be rounded

• The number of people in your family
• Amount of pencils in a pencil case
• A dozen = 12

Rounding:
When Rounding, you will round up when the digit is five or up and you will round down if digit is 4 and under
HOWEVER: if the digit 5 is the last digit, then you would round to the nearest even number (zeros after the five follow this rule too)

Math and Significant Figures
When adding or subtracting significant figures, round the answer to the least amount of decimal digits shown in the question.

Example:

  23000

  15600

+33120

  71720

However, because there are certain digits that you are unsure of, 71720 is NOT your final answer. The question would actually be something like this:

  23???

  156??

+3312?

  71???

Since we know that hundreds place is a 7, we would round our answer up and so the answer to this addition question, with consideration of significant figures would be: 72000

Multiply and Dividing

When multiply or dividing, we round the answer to the least amount of significant figures present in the question.

Example:

135x1 = 100

This is because 1 only has one significant digit and so when the product is 135, we round it so it would only have 1 significant digit (the hundreds) and so it becomes 100.

23.56/1.12 = 21.0

Since 1.12 only has three significant figures, the answer to this question will only have 3 significant figures as well.