4.5 Mass Ratio Calculations

​4.5 Mass Ratio Calculations

​The mass ratio gives the mass of an element that is found in combination with another element.

Multiple Choice

What distinguishes elements from compounds?

• Elements are made of one type of atom; compounds are made of two or more types of atoms.

• Elements are made of two or more types of atoms; compounds are made of one type of atom.

• Elements are always gases; compounds are always solids.

• Elements are always metals; compounds are always non-metals.

One of the fundamental laws of chemistry deals with the fact that we cannot (using chemical means) create or destroy matter.  When a reaction is run, the number of atoms of each specific type must be the same on both sides of the equation.  This is known as the Law of Conservation of Matter.
For some materials, it turns out that one element can combine with a second element in more than one ratio.  Think CO and CO_2.

Carrying out mass ratio calculations helped establish the law of multiple proportions, which we learned about in CK 4.3.

Multiple proportions = two or more elements, different compounds, simple ratios between them.

​Law of Definite Proportions CK 4.4

• A chemical compound always contains the same elements in exactly the same ratio by mass, no matter where it comes from or how it’s made.

• Example: Water (H2​O) is always about 89% oxygen and 11% hydrogen by mass.

Law of Multiple Proportions CK 4.3

• Two elements that can form more than one compound together.

• It says that if you compare the masses of one element that combine with a fixed mass of the other, those masses are in simple whole-number ratios.

• Example: Carbon and oxygen can make carbon monoxide (CO) and carbon dioxide (CO₂​). If you keep the mass of carbon the same, the mass of oxygen in CO_2​ is exactly twice that in CO.

​Carbon forms two different compounds with oxygen:

• In carbon monoxide (CO), 12 grams of carbon combine with 16 grams of oxygen.

• In carbon dioxide (CO₂), 12 grams of carbon combine with 32 grams of oxygen.

  Find the ratio of oxygen that combines with the same amount of carbon (12 g):

  • In CO: 16 g O

  • In CO₂: 32 g O

• Write the ratio of oxygen in CO₂ to CO:

  Ratio=32:16​=2:1

• What does this mean?

  • For the same amount of carbon, the mass of oxygen in CO₂ is exactly twice that in CO.

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Conclusion:
This shows the law of multiple proportions:
When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios (here, 2:1).

Multiple Choice

What is the definition of multiple proportions?

• Principle in chemistry that states the ratio of the masses of elements in compounds can be expressed in large whole numbers.

• Principle in chemistry that states the ratio of the masses of elements in compounds can be expressed in small whole numbers

• Principle in chemistry that states the ratio of the masses of elements in compounds can be expressed in decimal numbers.

• Principle in chemistry that states the ratio of the masses of elements in compounds can be expressed in fractions.

Multiple Choice

Explain the concept of stoichiometry in relation to multiple proportions.

• Stoichiometry is the study of the physical and chemical properties of elements and compounds.

• Stoichiometry is the process of balancing chemical equations.

• Stoichiometry is the measurement of the volume of gases involved in a chemical reaction.

• Stoichiometry is the calculation of the quantities of reactants and products in a chemical reaction.

Multiple Choice

In the example of peroxide (H₂O₂) and water (H₂O), what is the ratio of oxygen mass in peroxide to water?

• 1:1

• 2:1

• 1:2

• 3:1

Multiple Choice

In carbon dioxide, how many grams of oxygen would there be if there are 24 grams of Carbon?

Remember that there is 2.66 g of oxygen for every gram of Carbon.

• 64 g

• 73 g

• 55 g

• 88 g

Multiple Choice

In carbon monoxide, there is always 1.33 g of Oxygen for 1 g of Carbon.

How many grams of Carbon are there in a 16.03 g sample of carbon monoxide?

Please round your answer to 3 sig figs.

• 7.25 g

• 6.34 g

• 6.88 g

• 7.23 g

• Definite proportions = one compound, fixed ratio. This law is sometimes referred to as the Law of Constant Composition.

• Key idea: No matter the sample size, the ratio of magnesium to oxygen in magnesium oxide is always the same. That’s the Law of Definite Proportions!

• Find the ratio of the first sample:

  • Mg:O = 12g : 8g simplifies to 12/8 = 3/2. So for every 3 grams of magnesium, there are 2 grams of oxygen.

• Apply the ratio to the second sample:

  • The second sample has 3 grams of magnesium.

  • Using the ratio,
    3 g Mg:2 g O
    
    
    

Key idea:
No matter the sample size, the ratio of magnesium to oxygen in magnesium oxide is always the same. That’s the Law of Definite Proportions!

​A sample of magnesium oxide (MgO) is found to contain 12 grams of magnesium and 8 grams of oxygen.
If you have another sample of magnesium oxide that contains 3 grams of magnesium, how many grams of oxygen should it contain?

​Find the ratio in the first sample:

• Magnesium : Oxygen = 12 g : 8 g = 12/8 = 3/2​
  So, for every 3 grams of magnesium, there are 2 grams of oxygen.

1. Apply the ratio to the second sample:

   • The second sample has 3 grams of magnesium.

   • Using the ratio,
     3 g Mg:2 g O

   • The second sample should contain 2 grams of oxygen

Multiple Choice

The law of definite proportions states that a given chemical compound always contains the same _ in the exact same _ by mass.

• elements, compounds

• elements, molecules

• elements, proportions

• proportions, elements

Multiple Choice

What does the Law of Definite Proportions state about the composition of elements in a compound?

• The composition varies with the amount of the compound.

• The composition remains the same regardless of the amount.

• The composition changes with temperature.

• The composition depends on the source of the compound.

Multiple Choice

What is the composition of water according to the law of definite proportions?

• 1 Hydrogen to 2 Oxygen

• 2 Hydrogen to 1 Oxygen

• 1 Hydrogen to 1 Oxygen

• 2 Oxygen to 1 Hydrogen

Fill in the Blanks

Multiple Choice

Water is 11.19% hydrogen and 88.81% oxygen by mass.

Lake Superior hold about 1.2 x 10¹⁶ kg of water.

What mass of hydrogen is in Lake Superior?

• 1.5 x 10¹⁶ kg

• 5.0 x 10¹⁵ kg

• 1.3 x 10¹⁵ kg

• 1.2 x 10¹⁶ kg

​Mass Ratio Calculations

Mass ratio calculations are all about comparing the masses of different parts in a chemical compound or reaction.

Here’s how it works:

• You find the mass of each element in a compound.

• Then, you compare those masses as a ratio.

Copper reacts with chlorine to form two compounds. Compound A consists of 4.08 g of copper for every 2.28 g of chlorine.  Compound B consists of 7.53 g of copper for every 8.40 g of chlorine. 
What is the lowest whole number mass ratio of copper that combines with a given mass of chlorine?

Step 1:  List the known quantities and plan the problem.

Compound A = 4.08 g Cu and 2.28 g Cl

Compound B = 7.53 g Cu and 8.40 g Cl

Apply the law of multiple proportions to the two compounds.  For each compound, find the grams of copper that combine with 1.00 g of chlorine by dividing the mass of copper by the mass of chlorine. 
Then, find the ratio of the masses of copper in the two compounds by dividing the larger value by the smaller value.

Compound A
Cu/Cl = 4.08 g/2.28g, if we assume 1.00 g of Cl, the Mass Ratio sets up like this: 4.08/2.228 = Cu/1.00, so let's cross multiply and solve.
4.08(1) = 2.228(Cu)
Cu = 1.79 g for every 1 g of Cl in Compound A

Compound B
Cu/Cl = 7.53 g/8.40 g, so if we assume 1.00 g of Cl, the Mass ration sets up this:
7.53/8.40 = Cu/1.00 and we cross multiply to solve.
7.53(1) = 8.40(Cu)
Cu = 0.896 for every 1 g of Cl in Compund B

Copper reacts with chlorine to form two compounds. 
Compound A consists of 4.08 g of copper for every 2.28 g of chlorine. 
Compound B consists of 7.53 g of copper for every 8.40 g of chlorine. 
What is the lowest whole number mass ratio of copper that combines with a given mass of chlorine?

Compare the masses of copper per gram of chlorine in the samples

1.79 g Cu (in Compound A) = 2.00 = 2:1
0.896 g Cu (in Compound B) 1

Think about your result.

The ratio is a small whole-number ratio.  For a given mass of chlorine, compound A contains twice the mass of copper as does compound B. This satisfies the Law of Multiple Proportions!

Multiple Choice

What are the laws of definite and multiple proportions and why are they important?

• They describe how elements combine in fixed ratios to form compounds.

• They explain the behavior of gases under pressure.

• They determine the color of compounds.

• They predict the melting points of elements.

Multiple Choice

What does mass best represent?

• The amount of space an object takes up

• The number of particle collisions on a container's walls

• The amount of matter in a substance

• The number of particles in a compound

Multiple Choice

Find the ratio  $$\frac{mass\ N}{mass\ O}$$  for compound A

• 5.34

• 1.75

• 0.734

• 1.19

Multiple Choice

Find the ratio of $$\frac{mass\ N}{mass\ O}$$  for compound B

• 1.75

• 1.36

• 1.14

• 0.876

Multiple Choice

Cmpd. A has a N:O mass ratio of 1.75

Cmpd. B has a N:O mass ratio of 0.876.

Which statement is true when comparing A to B?

• A has more Oxygen atoms than B

• There is more Nitrogen than Oxygen in compound B

• There is twice the amount of Nitrogen per Oxygen in A than in B

• Compound A is twice as massive as compound B