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Chemical Conversions and Problems
Contents: Â Temperature Conversions: pg. 1 Â Basic Unit Conversions: pg. 1- Â Chemical Quantity Conversions: pg. 3- Â Density: pg. 5- Â Percent Composition & Empirical & Molecular Formula Determination: pg. 6- Â Stoichiometric Calculations: pg. 8- Â Limiting Reagents: pg. 9- Â Percent Yield: pg. 11
Temperature Conversions To convert between temperature in Kelvins (TK) and degrees Celsius (T°C): TK = T°C + 273.15 or T°C = TK â 273.
To convert between temperature in degrees Celsius (T°C) and degrees Fahrenheit (T°F): T°C = T°F â 32 or T°F = 1.80(T°C) + 32
Example 1: What is the boiling point of water, 100°C, in degrees Fahrenheit?
T°F = 1.80(T°C) + 32 & T°C = 100 T°F = 1.80(100) + 32 T°F =180 + 32 T°F = 212
Basic Unit Conversions To do a basic conversion from one unit to another:
- Start with the original number you are given. In example 2, the original number given is 34 minutes.
- Multiply/divide that original number by a known relationship between that original unit and the unit you want to end up with. In example 2, the known relationship is 60 seconds equals 1 minute.
How you determine whether to put the 60 s on the top and the 1 min on the bottom or put the 1 min on the top and the 60 s on the bottom depends on which unit you are starting with and which unit you need to end up with. You want to be able to cancel out the original unit leaving the ending unit on top; put the unit you began with, min, on bottom and the unit you want to end up with, s, on top. In Example 2, you want to be able to cancel out the min, so you put min on bottom, and you want to end up with s, so put s on top. If you were to do a problem converting from seconds to minutes, however, you would put min on top and s on the bottom as in the Example 3.
Example 2: How many seconds are in 34 minutes?
Example 3: How many minutes are in 25 seconds?
Often in conversion problems, you will have to use more than one conversion factor (known relationship) as in Example 4. Just take these multi-step conversions one step at a time, canceling out all units except the one you want to end up with.
Example 4: How many centimeters are in 10 miles?
Known relationships: 1 mi = 5280 ft, 1 ft = 12 in, and 1 in = 2.54 cm
(If you need to review how to put numbers in scientific notation, see our handout Exponents, Radicals, and Scientific Notation .)
Another type of common conversion problem deals with conversions between some unit and a prefix of that unit such as a conversion from meters to millimeters. The following table provides a list of some widely used prefixes. For example, 1 gigameter (Gm) = 1,000,000,000 meters (m) or 10^9 m, and 1 microinch (Îźm) = 0.000001 in or 10 -6^ in.
Prefix Value Name Symbol Numerical Exponential giga G 1,000,000,000 109 mega M 1,000,000 106 kilo k 1,000 10 3 hector h 100 10 2 deka da 10 101 deci d 0.1 10 - centi c 0.01 10 - milli m 0.001 10 - micro Îź 0.000001 10 - nano n 0.000000001 10 - pico p 0.000000000001 10 - femto f 0.000000000000001 10 -
Example 8: What is the molar mass of H 2 SO 4?
H: 2(1 g/mol) = 2 g/mol S: 1(32 g/mol) = 32 g/mol O: 4(16 g/mol) = 64 g/mol 98 g/mol
In Example 8, H 2 SO 4 was made up of 4 O atoms, so the atomic mass of O was multiplied by
- Likewise, the atomic masses of H and S were multiplied by 2 and 1 since there were 2 atoms of H and 1 atom of S. Once you have solved for the molar mass of a molecule, you can use it to convert between moles and grams in a conversion problem set up just like basic unit conversions. This is possible because the molar mass of any compound equals 1 mole of that compound.
Example 9: How many grams are in 50. moles of H 2 SO 4?
Known relationship: 1 mol = the molar mass of H 2 SO 4 = 98 g
Once you know how to convert between moles and grams and between moles and numbers of atoms or molecules, you can perform conversions between grams and number of atoms or molecules by putting together the two known relationships between moles and grams and between moles and number of atoms or molecules.
Example 10: How many kilograms of Ne gas are there if there are 104 atoms of Ne?
Known relationships: 1 mol = 6.022 x 10^23 atoms, 1 mol = molar mass of Ne = 20g, 1 kg = 10 3 g (or 1000 g)
In some problems, you may have to convert between moles or grams of some molecules and atoms of an element in that molecule. For example, you might be given moles or grams of C 2 H6 and be asked to find the number of H atoms in that sample. In problems like this, you can use the fact that there are 6 atoms of H in 1 mlcl of C 2 H6 as a conversion factor.
Example 11: How many atoms of H are there in 124 g of NH 3?
Known relationships: 1 mol = molar mass of NH 3 = 17 g, 1 mol = 6.022 x 10^23 mlcls, 1 mlcl NH 3 = 3 atoms H
Density The density of a substance is the ratio of that substanceâs mass to its volume. The higher the density of a substance, the less space it takes to fill that space with some amount of the substance. For example, imagine you have 100 pounds of lead and 100 pounds of cotton balls. It would take a space with a much larger volume to accommodate 100 pounds of cotton balls than lead. This is because lead is much denser than cotton balls. To determine the density of a substance, divide the amount of that substance by the volume it takes to accommodate that amount. If you know the density, you can solve for the volume a certain mass will occupy or you can solve for the mass a certain volume will accommodate.
Density = Mass or Volume = Mass or Mass = Volume x Density Volume Density
Example 12: What is the density of 908 g of cadmium metal that occupies 105 mL?
Density = Mass = 908 g = 8.65 g/mL Volume 105 mL
The important thing is to first make sure you are dealing with the correct units before solving for density. Mass should be in grams, and volume should either be in mL or cm^3 since 1 mL = 1 cm^3. This means density should always be in g/mL or g/cm^3. You may have to do some basic unit converting first.
Example 13: What mass of Ne gas with a density of 0.00090 g/mL occupies a volume of 70 ounces?
Known relationships: 32 oz = 1 qt (quart), 1.0567 qt = 1 L, and 1 mL = 10-3^ L (or 0.001 L)
Sometimes density can be incorporated into other types of problems, for example, chemical quantity conversions. In Example 14, you must find the mass from the density and volume before the number of moles can be found.
Example 14: How many moles are there in a sample of titanium metal, density 4. g/mL, that can occupy a volume of 25.0 mL?
Known relationships: 1 mol = molar mass of Ti = 48 g
You can find the empirical formula of a compound if you know the percent composition of that compound. To do so:
- Designate the total amount of sample as 100 g so that the percentage of each element will equal the amount in grams of each element.
- Convert grams of each element into moles.
- Divide each number of moles by the smallest number of moles. In example 16, the smallest number of moles is Cr, so moles of K, moles of Cr, and moles of O will each be divided by moles of Cr.
- Put the numbers obtained from dividing by the smallest number into a ratio.
- Multiply that ratio by the smallest whole number that will give a ratio of close to whole numbers.
Example 16: What is the empirical formula of a compound containing 26.57% K, 35.36% Cr, and 38.07% O?
The molecular formula of a compound can be determined if you know the empirical formula and the molar mass of the compound. To do so:
- Convert moles of each element into grams.
- Add the grams of each element together to get a total empirical formula weight.
- Divide the molar mass by the empirical formula weight.
- Multiply the subscripts of each element by the number obtained from step 3.
Example 17: What is the molecular formula of ethylene glycol if its empirical formula is CH 3 O and its molar mass is 62.1 g?
Stoichiometric Calculations Stoichiometric calculations are calculations that involve the relationship between the quantity of products and of reactants of a chemical equation. With a balanced chemical equation, you can convert between moles of reactants and products using the coefficients. To convert from the amount of one participant in a reaction to another participant in a reaction:
- Make sure the reaction given is balanced (for balancing review, see our Balancing Chemical Equations handout).
- Convert the amount of the compound given to moles if it is not already.
- Convert to the desired compound by doing a mole to mole ratio using the coefficients of the two compounds youâre converting between. Treat the mole to mole ratio like a conversion factor (known relationship) from basic unit conversion problems. In example 14, 4 mol NH 3 = 3 mol O 2 , 4 mol NH 3 = 2 mol N 2 , 4 mol NH 3 = 6 mol H 2 O, 3 mol O 2 = 2 mol N 2 , 3 mol O 2 = 6 mol H 2 O, and 2 mol N 2 = 6 mol H 2 O are all conversion factors that can be used to convert between compounds in the reaction 4NH 3 + 3O 2 â 2N 2 + 6H 2 O. Remember to put the compound you want to end up with on the top and the compound you started with on the bottom of your ratio so that the compound you started with can be canceled out just like a unit.
- If the problem asks for the end compound to be in some unit other than moles, do that conversion.
Example 18: How many moles of H 2 O are produced if 40 moles of NH 3 are used?
Example 19: If 16 grams of O 2 are used in the following reaction, how many grams of C2 H6 are used? 2C 2 H 6 + 7O 2 â 4CO 2 + 6H 2 O
Known relationships: 2 mol C 2 H6 = 7 mol O 2 , 1 mol = molar mass of C 2 H6 = 30 g, and 1 mol = molar mass of O 2 = 32 g
Example 20: A solution containing 2.00 g of C 5 H 11 OH was added to a solution containing 2.00 g of O 2. Which reagent is the limiting reagent? How much of the reagent that is not limiting will be left over?
Since the amount of C 5 H11 OH that will react with all the O 2 , 1.67 x 10-2^ mol is less than the amount of C 5 H11 OH actually used, 2.27 x 10-2^ mol, there will be leftover C 5 H 11 OH, and O 2 will be the limiting reagent. This could also be determined if you determined how much O 2 would be needed to react with all the C 5 H11 OH. That amount would be more than the available amount of O 2.
Since O 2 is the limiting reagent, all of it will be consumed, and there will be leftover C 5 H 11 OH.
Example 21: How many grams of Ca 3 (PO 4 ) 2 and grams of KCl can be produced by mixing 5.00 g of CaCl 2 with 8.00 g of K 3 PO 4?
3CaCl 2 + 2 K 3 PO 4 â Ca 3 (PO 4 ) 2 + 6KCl
CaCl 2 is the limiting reagent since there is not enough of it for all the K 3 PO 4 being used to be consumed. 0.0566 mol CaCl 2 would be needed for all the K 3 PO 4 to be consumed, but we only have 0.0455 mol CaCl 2.
Percent Yield The theoretical yield of a reaction is the amount of product that was calculated to be produced. Example 21 shows how to calculate the theoretical yield of a product or products. Percent yield is the percentage of the theoretical yield that was actually obtained when the chemical reaction was performed. To calculate the percent yield, use the following formula:
Percent Yield = Actual Yield x 100% Theoretical Yield
Example 22: The theoretical yield of H 2 O in the combustion of CH 4 was calculated to be 55 g, but when the reaction was actually performed, the amount of H 2 O yielded was 51.7 g. What is the percent yield?
Percent Yield = Actual Yield x 100% Theoretical Yield
