Formula of a Hydrate Lab (by Bev)

What is a hydrate?

A hydrate is a molecule that has a definite number of water molecules in its crystal structure.  The crystals seem like a dry solid, but they can be decomposed into an anhydrous salt (without water) and water when heated.  A colour change often happens when the hydrate is decomposed

     ex. calcium chloride hexahydrate: CaCl2*6H2O
                6mol H2O are chemically combined to 1mol CaCl2

What are hydrates used for?

Grains of anhydrous CaCl2 can be used to in damp rooms to absorb the moisture out of he air, thereby forming a hydrate.

Lab 4C: Formula of a hydrate


The mass of a dry crucible was found on a centigram balance & the hydrate used in this experiment, MgSO4*7H2O, was heated in a dry crucible over a Bunsen burner and weighed as well.  Using this information, and mass of the anhydrous salt & water was calculated.

%H2O in hydrate: mass of H2O/mass of hydrate = 2.23g/5.02g x 100% = 44.4%

# of mol of MgSO4 in hydrate:
mass after heating-mass of empty crucible = mass of anhydrous salt
28.90g-26.11g = 2.79g

1mol anhydrous salt = mass anhydrous salt x mol/120.4g = 2.79g x mol/120.4g = 0.02mol

mass of hydrate = mass of crucible & hydrate - mass of crucible = 31.13g-26.11g = 5.02g

mass of H2O = mass hydrate - mass of salt = 5.02g - 2.79g = 2.23g

1mol H2O =  mass of H2O x mol/18g = 2.23g x mol/18g = 0.1239mol

mass of salt/1mol salt = 2.79g/120.4g/mol = 0.023mol

mol H2O/mol salt = ratio of H2O to anhydrous salt = 0.1239mol/0.23mol = .539

Empirical and Molecular Formula - Isabelle Cheng

Isabelle Cheng

Block 2-2 Chemistry 11

Ms. Chen

Empirical and Molecular Formula

Empirical - smallest whole number ratio of atoms which represent the molecular composition of a species. In other words “species” means elements. It is where a number the numbers can be divided by a certain number. For example, C8H16 it would be reduced down to C2H4. That is the lowest terms in this case.

Here is one example of doing an Empirical Formula question.

Say that you have 54.09% of Ca, 43.1% of O and 2.73% H. What is the empirical formula?

The first step:

Separate them into elements. 

Ca

O

H

The second step:

Now you want to convert grams into moles. This means you have to write the mass on the side now. 

Ca =  54.09

O = 43.1

H = 2.73

The third step:

Now you have to multiply it by one mole and divide it by the mass of the element (species).

Ca = 54.09/40.1 = 1.349

O = 43.16/ 16= 2.699

H = 2.73/1 = 2.73

   

The fourth step:

Now divide both by the smallest molar amount.

Ca = 1.349/1.349 = 1

O = 2.699/1.352 = 2

H = 2.73/1.352 = 2

The fifth step:

Now put them together:

Ca(OH)2 which is Calcium Hydroxide!

Now moving onto molecular formula!

Molecular Formula - it is a multiple of many empirical formula and has the real number of atoms that combine to form a molecule!

The formula for this is:

Let N = the WHOLE NUMBER MULTIPLE OF THE EMPIRICAL MASS:

multiple N =       Molar mass

Empirical mass

One example of the Molecular Formula is this:

A hydrocarbon is 84.25% carbon and 15.75% hydrogen and has a molecular weight of 114. What is the molecular formula?

C = 84.25/12 = 7.021

H = 15.75/1 = 15.75

Then you go:

7.021/7.021 = 1

15.78/7.021 = 2.25

Next step:

C = 1 X 4 = 4

H = 2.25 X 4 = 9

Now you know that answer!

It is.....: 

4 C = 48

9 H = 9

Percent Composition (by Bev)

Percent composition is the percentage by mass of a species in a chemical formula

ex. 1: percent composition of AgOH (assume there is 1mol)

total MM (molar mass): 107.9 + 19.0 + 1.0 = 127.9
MM of Ag: 107.9 (same numerical value as the atomic mass)
MM of O: 16.0
MM of H: 1.0

% Ag: 107.9g/mol / 127.9g/mol x 100% = 84%
% O: 19.0g/mol / 127.9g/mol x 100% = 15%
% H: 1.0g/mol / 127.9g/mol x 100% = 1%

Check: 84% + 15% + 1% = 100%*

*the percentages should add up to or very close to 100%

ex. 2: percent composition of CO2 (assume there is 1mol)

total MM: 12.0 + 16.0(2) = 44.0g/mol
MM of C: 12.0g/mol
MM of O: 32.0g/mol

% C: 12.0g/mol / 44.0g/mol x 100% = 27.3%
% O: 32.0g/mol / 44.0g/mol x 100% = 72.2%

Check: 27.3% + 72.2% = 100%

ex. 5: percent composition of  O in Cu2CO3

total MM: 123.5g/mol
MM of O: 48.0g/mol

% O: 48.0g/mol / 123.5g/mol x 100% = 39%

ex. 4: a compound contains 5.1 g of Cl, 22.0g of c, an unknown mass of O, & the total mass is 44.1g.  Calculate % composition.

Mass of O: 44.1g-5.1g-22.0g = 17.0g

% Cl: 5.1g / 44.1 g x 100% = 11.6%
% O: 17.0g / 44.1g x 100% = 38.6%
% C: 22.0g / 44.1g x 100% = 49.9%

Check: 11.6% + 38.6 % + 49.9% = 100.1 %

Links to get you started:
http://chemistry.about.com/od/workedchemistryproblems/a/mass-percent-worked-problem.htm
http://www.ausetute.com.au/percentc.html

http://www.youtube.com/watch?v=qzUMKWhWKm8 (THE BEST LINK!)
http://www.youtube.com/watch?v=CSuZVJ8TA40&feature=related

Converting THE MOLE

Mole conversions. They may seem hard, but they're just like unit conversions.

 Things to remember:
-Avogadro's number (SUPER IMPORTANT!!)
-how to calculate molar mass
-the "magic 1" rule
-SIG. FIGS!

OK. Let's start converting!!

*For these conversions, we'll need to use Avogadro's number: 6.022x10^23*

1. particles-->moles
4.24x10^24 C particle-->moles

4.24x10^24particles x                1 mole         = 7.04 moles of C
                                   6.022x10^23 particles

2. moles-->particles
1.34 moles moles CO2-->molecules

1.34moles x 6.022x10^23 molecules = 8.07x10^23 molecules of CO2
                               1 mole

These conversions do not require Avogadro's number. They instead, require a periodic table. Remember how to calculate molar mass!

3. moles-->grams
6.5 moles of C-->grams                                        *Use molar mass of C=12.0g/mol

6.5 moles x 12.0g/mol = 78 grams of C
                      1 mole

4. grams-->moles
8.2 grams of MgCl2-->moles                                *Use molar mass of MgCl2=95.3g/mol

8.2 grams x 1 mole = 0.086 moles of MgCl2
                   95.3 g


YAY!! So now, you should know how to convert from moles<->particles and moles<-> grams.
Just for some extra practice, here are a few questions:

1. How many moles of O2 are in 5.53x10^41 molecules of Ag?

2. How many atoms are in 2 moles of C?

3. How many moles are in 94.0g of Pb?

4. Calculate the mass (g) of 7.42x10^12 atoms of C.





And, lastly, here's a video to summarize the converting stuffs. Don't get distracted by the music!
Teehee.

 

Written by Jialynn.

Chapter 4 - Atomic Mass and Avogadro's number - Isabelle Cheng - November 17, 2010

Isabelle Cheng

November 17, 2010

Block 2-2 Chemistry 11

Ms. Chen

Chapter 4 - Avogadro’s Theory and Atomic Masses

The mole is the amount of carbon atoms in 12 grams of carbon. On the other hand, a molar mass is the mass of one mole. For example, a dozen means 12 and so, a mole means 12 grams. Using the periodic table is very necessary for solving these problems. Some examples from using the periodic table is that Iron has an atomic mass of 55.8. Then right away you know that the molar mass of the element is also 55.8g. 

Another example:

AgNO3 - 1 Ag = 1 x 107.9 = 107.9, 1 N = 1 x 14.0, and 3 O = 3 X 16.0 = 48.0 and all of it together is 169.9 grams which means that the molar mass is 169.9 grams.

We also need to include unit conversions in these problems. 

For example, we need to use this equation:

1mol___     molar mass of X

molar mass of X           1mol

Avogadro’s number:

- 6.022 X 10 ^ 23

- large number 

Experiment 2E: Determining Aluminum Foil Thickness by: Mandy

Lab 2E : Determining Aluminum Foil Thickness

Today, we did experiment 2E. This lab is an indirect measurement of the thickness of a piece of aluminum foil.

There were two formulas involved in this experiment:

Volume of a rectangular solid = length X width X height

Density of a substance = mass / volume

We first took three aluminum foils (15 X15)cm. since the measurement of those square aluminum foils might be off by a little bit, we need to measure the length of two widths and take their average; same as the lengths. Then, we took each aluminum foil to the centigram balance for the measurement of mass.  Now that we knew the lengths, widths, mass, and density (2.70 g/cm^3), the height can be calculated.

In this experiment, accuracy and precision are very important.

An ACCURATE measurement is a measurement that is close to the accepted values.

A PRECISE measurement id a reproducible measurement, thus, more precise the measurement is, the more significant digits it has.

Take the average thickness of those three aluminum foils, and calculate experimental error.

Experimental Error is calculated using this formula:

Experimental Error = abs( your measurement – accepted value) /( accepted value) X 100%

In this case, the accepted value is: 1.55*10^-6 cm



household aluminum foil



a little extra information about aluminum foil :) 

        The shiny side is slightly better reflector of heat. To keep things cold, put the shiny side on the outside [that will reflect incoming heat]. To keep things warm, face the shiny side inward toward the hot food [to reflect the heat that is trying to escape back into the food].

THANK YOU for checking out our blog :D

So Like, How Much More "Dense" Can You Get?

Things to remember:

1cm^3 of water=1mL

Density of water=1.0g/mL OR 1000g/L

The formula for density is Density=mass/volume

Density Units

for solids--> g/cm^3

for liquids--> g/mL

Some concepts to think about:

If the density of an object is greater than the density of a liquid, the object will sink
If the density of an object is less than the density of a liquid, the object will float.


Try applying the formula to these practice problems. Remember, D=m/v.

1. An iron bar has a mass of 1225g and a volume of 1.2L. What is the iron bar's density?

2. In a balloon, helium occupies 3.8L with a mass of 4.0g. What is the density of helium?

3. An iron bar has a mass of 250g. If iron's density is 7.86x10^3 g/L, what volume does the bar occupy?

4. A block of beeswax has a volume of 210.0mL and a density of 961g/L. What is the mass of the block?

Fun Fact!!

The density of one brain tissue is 1.05g/mL, which is almost equivalent to the density of water!
How "dense" is that??!?! xD


Written by Jialynn