Today in chem class we conducted an experiment in which we were to determine whether or not stoichiometry accurately predicts the mass of products produced in chemical reactions. The procedure was very similar to the Mole Ratio Lab we did in November, 2011.
*The balanced chemical equation for this reaction is: Sr(NO3)2 + CuSO4 --> SrSO4 + Cu(NO3)2
*Therefore, since 2.00g of Strontiumnitrate completely reacts, the accepted value of grams of Strontium sulphate that is produced is: 2.00g x 1mol/211.6g x 1/1 x 183.7g/1mol = 1.736294896g or 1.74g
STRONTIUM NITRATE:
COPPER II SULPHATE:
In this experiment, 2.00g of Strontium nitrate (dissolved in water) was reacted with 3.00g of excess Copper II sulphate (dissolved in water). A precipitate was formed (Strontium sulphate), as well as a bright blue solution (Copper II nitrate). We separated the precipate by filtering, removed the water by drying, and found the mass by weighing.
The mass of Copper II sulphate used was 3.00g.The mass of Strontium nitrate used was 2.00g. The mass of the filter paper was 0.84g.The mass of precipitate and filter paper was 2.62g. And the mass of the precipitate (combination - filter paper) was 1.78g.
According to our observations, 0.00969 moles of precipitate were formed. Calculations: 1.78g x 1mol/183.7g = 0.00969mol (183.7g is the molar mass of Strontium sulfate)
The percent error in this experiment was 2.30% error. We are very pleased with these results.
Error exists in this experiment for a number of reasons. It is possible that our beakers were not entirely clean and contained other substances.There could have been miscalculations in measuring the amount of water and amount of copper II sulphate and strontium nitrate (it may not have been exactly 3.00g of Copper II sulphate and 2.00g of Strontium nitrate). It's possible that the filter paper was not dry enough (causing it to have a greater mass than it should). It's also possible that some precipitate may have transferred to our filter paper from other groups when they were shifted around in the drying oven.
Today's lesson was simply an extension of last class's material. Mass to mass problems involve one additional conversion:
After writing the balanced chemical formula, find the mass of the first substance. Use its molar mass to convert it to mole units. Then, use the coefficients of the balanced equation (remember: what you need over what you have). Finally, use the molar mass of the new substance to determine the mass.
EXAMPLES:
*Chlorine reacts with 0.29g of sodium iodide. What is the mass of sodium chloride that is produced?
*When water reacts with 0.38g of lithium nitride, what is the mass of lithium hydroxide produced?
*How many grams of O2 are produced from the decomposition of 3.0g of potassium chlorate?
In today's class we learned how to convert from moles to mass and mass to moles, which is an extension of what we learned last class.
Examples:
*How many moles of Oxygen gas are required to produce 9.8 mol of Carbon dioxide (decomposition)?
1. CO2--> C + O2
9.8 mol x 1/1= 9.8 mol
*When Magnesium chloride reacts with Potassium oxide to form Magnesium oxide and Potassium chloride, how many moles of Magnesium chloride are needed to produce 18.4 mol of Potassium chloride?
2. MgCL2 + K2O -> MgO + 2KCL
18.4 mol x 1/2 = 9.20 mol
In some cases you will be given a quantity that is in moles and the question will ask you to determine the mass.
*How many grams of Calcium chloride are needed to produce 8.34 mol of pure Calcium?
CaCl2-> Ca + Cl2
8.34mol x 1/1 x 111.1g/1mol= 926.574g or 927g
*How many grams of Calcium chloride are needed to produce 67.43 mol of pure Chloride?
67.43 mol x 1/1 x 111.1g/1 mol= 7491.473g or 7491g
Coefficients in balanced equations tell us the number of moles reacted or produced. They can also be used as conversion factors. This brings us to the golden rule of mole to mole conversion, what you need over what you have.
How many moles of Calcium Phosphide are required to produce 2.3 mol of pure Calcium
2Ca3P2 -> 6Ca + P4
2.3 mol x 6/2 = 3 mol of Calcium Phosphoide
When 0.7 mol of Scandium react with Chromium (III) Fluoride how many mole of Chromium should be produced?
Sc + CrF3 = Cr + ScF3
0.7 mol x 1/1 = 0.7 mol. Of Chromium
For more explanation on mole to mole conversions check this video out!
Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions. It is a generalization of mole conversions to chemical reactions. The foundation of stoichiometry is understanding the 6 types of chemical reactions. The 6 types of reactions are; synthesis (formation), decomposition, single replacement (SR), double replacement (DR), neutralization, and combustion.
Synthesis
Synthesis reactions combine two elements to make a compound
- A+B -> AB
Examples:
4Fe + P4 -> 4FeP
Here is a helpful video for synthesis reaction:
Decompostion
Decomposition is the reverse of Synthesis. Always assume that compounds decompose into elements during decomposition.
- AB -> A+B
Examples:
Ca(NO2)2 -> Ca + 2(NO2)
Here is a helpful video for decomposition reactions:
Single Replacement (SR)
In single replacement reactions the metals switch places.
- A + BC -> B + AC
Here is a helpful video on single replacement reactions.
Examples:
Sr + 2LiF -> 2Li + SrF2
Double Replacement (DR)
In double replacement reactions the non-metals in each compound switch places.
- AB + CD -> AD + BC
Examples:
BeI2 + Na2(SO3) -> Be(SO3) + 2NaI
Here is a helpful video on double replacement reactions:
Neutralization
A neutralization reaction is a reaction between an acid and a base
Examples:
HCl + NaOH -> NaCl + HOH
Here is a helpful video on neutralization reactions:
Combustion
A combustion reaction is a reaction of something (usually hydrocarbon) with air. Hydrocarbon combustion always produces CO2 + H2O
For our final lesson in this unit, we spent about ten minutes learning molecular formulas. Then, we spent the rest of the class studying for our test on Thursday. This topic is really quite simple if you already understand the previous lesson. Let's get to it!
Molecular formulas tell us exactly how much of each atom are in a compound. If we know the empirical formula and the molar mass, we can easily determine the molecular formula. Here is yet another chart to help us write a molecular formula:
That really is all there is to it! Told you it was easy! Now for some examples:
EXAMPLES: -The empirical formula for a substance is CH2O and its molar mass is 60.0g/mol. Determine the molecular formula.
Simply find the molar mass of the empirical formula using your periodic table. You should arrive at 30.0g/mol. Then, divide 60 by 30, and you should arrive at 2. So, multiply all of the subscripts in the empirical formula by 2 to arrive at the molecular formula, C2H4O2.
-The empirical formula for a compound is C2H6O and the molar mass is 138g/mol. Determine the molecular formula.
Simply find the molar mass of the empirical formula using your periodic table. You should arrive at 46g/mol. Then, divide 138 by 46, and you should arrive at 3. So, multiply all of the subscripts in the empirical formula by 3 to arrive at the molecular formula, C6H18O3.
We're back from Christmas break and we're starting the New Year with some chemistry. Yay!!
Empirical formulas are the simplest formula of a compound. In fact, they show only the simplest ratios rather than the actual number of atoms (that would be a molecular formula). The empirical formula for Chlorine gas is simply Cl. This means that Dinitrogen tetraoxide would not be written as N2O4. Here are some examples of empirical formulas:
To determine the empirical formula we need to know the ratio of each element. To determine the ratio, fill in the table below for each problem:
If the smallest ratio is in decimals, you need to multiply everything by a common number. For example, multiply 0.5 by 2. Multiply 0.33 or 0.66 by 3. Multiply 0.25 or 0.75 by 4. Multiply 0.2, 0.4, 0.6, and 0.8 by 5.
EXAMPLES: -A sample of an unknown compound is found to contain 8.4g of 'C', 2.1g of 'H' and 5.6g of 'O'. Determine the empirical formula. **Note that some information filled out for each atom is not initially determined. ATOM: C -----MASS: 8.4 -----MOLAR MASS: 12.0 -----MOLES: 0.70 -----MOLE / SMALLEST MOLE: 2 -----RATIO: 2
Divide by the smallest mole, 0.35! 0.70/0.35 = 2 2.1/0.35 = 6 0.35/0.35 = 1
Therefore, the empirical formula is: C2H6O
-An unknown compound that has a total mass of 0.888g is made up of Carbon, Hydrogen and Oxygen. The mass of Carbon is 0.576g while the Hydrogen totals 0.120g. Determine the empirical formula. **Note that some information filled out for each atom is not initially determined. 0.888 - (0.576 + 0.120) = 0.192g
Divide by the smallest mole, 0.012! 0.048/0.012 = 4 0.12/0.012 = 10 0.012/0.012 = 1
Therefore, the empirical formula is: C4H10O1
-Determine the empirical formula of a compound that is 50.5% C, 5.26% H and 44.2% N. **Note that some information filled out for each atom is not initially determined. We can assume that the percentages actually represent the mass.
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