Scientific Understanding

**What is Effusion?**

Effusion and Diffusion are very similar. Diffusion is the movement of particles or molecules across one or more barriers with tiny holes in it. Effusion is a simplified type of Diffusion. That is, effusion is the movement of particles or molecules through only one hole between only two containers. Here is a video demonstration. Regardless, you do not really have to know any of that to succeed in this section. The equation for how fast (the rate) that effusion happens is below. The equation is what is important.

Rate of gas 1 = | Square root of Molar Mass of gas 2 |

Rate of gas 2 | Square root of Molar Mass of gas 1 |

Or short hand

Rate of gas 1 = | Sqrt MM gas 2 |

Rate of gas 2 | Sqrt MM gas 1 |

Sqrt stands for SQUARE ROOT OF. MM stands for MOLAR MASS OF.This formula says that the rate of effusion or diffusion of a gas depends on the square root of its molar mass. That suggests that lighter gas effuse or diffuse slightly faster. These are not absolute rates. Meaning they are not always the same rate for a particular gas in any conditions. Rather they are relative rates. They are only a comparison of these two particular gases at these particular conditions. It is like saying a Lamborghini travels twice as fast as a Ford. In that comparison of cars I did not actually say how fast each of the cars were going. Instead, I compared them to one another. This formula is doing the same thing to gasses. If you try the numbers in the formula it says that He gas travels or effuses at twice the rate of CH_{4} gas. We will try this exact problem later as a demonstrated example. Be careful with this formula. Many people are fooled into writing this formula incorrectly or plugging in the numbers in the wrong spot because how the formula is written does not match how they are thinking of it. This will become clearer in the demonstrated examples.

**Examples**: Solve for the following effusion problems. Use this periodic table link to look up the molar masses.

If the rate of effusion of H_{2} gas is 15. What is the rate of effusion of N_{2} gas?

Answer: 3.96

If you have an unknown gas that effuses at a rate 2 times that of F_{2} what is the molar mass of that gas?

Answer: 9.5 g/mol

How much faster is the rate of effusion of O_{2} gas to that of CO_{2} gas?

Answer: 1.17 times faster

**VIDEO Effusion Demonstrated Example 1**: If the rate of effusion of Ne gas is 5. What is the rate of effusion of Cl_{2} gas? Use this periodic table link to look up the molar masses.

**Step 1:**

What information are we given?

Answer:

Rate of Ne = 5

(Look up) Molar Mass Ne = 20 g/mol

(Look up) Molar Mass Cl_{2} = 71 g/mol

**Step 2:**

What does the question ask for?

Answer: Rate of Cl_{2} = ?

**Step 3:**

How do we set up the problem?

Answer: Start with the equation

Rate of Ne = | Sqrt MM Cl_{2} |

Rate of Cl_{2} |
Sqrt MM Ne |

**Step 4:**

What can we fill in for the equation?

Answer: The information we are given (red).

5 = | Sqrt 71 |

Rate of Cl_{2} |
Sqrt 20 |

**Step 5:**

How do we rearrange the equation?

Answer: Multiply both sides by Rate of Cl_{2}(red).

Rate of Cl_{2 }(5) = |
Sqrt 71 * Rate of Cl_{2} |

Rate of Cl_{2} |
Sqrt 20 |

**Step 6:**

Cross out Rate of Cl_{2} on the left side.

Rate of Cl_{2}(5) = |
Sqrt 71 * Rate of Cl_{2} |

Rate of Cl_{2} |
Sqrt 20 |

**Step 7:**

Simplify

(5) = | Sqrt 71 * Rate of Cl_{2} |

Sqrt 20 |

**Step 8:**

How do we rearrange the equation?

Answer: Multiply both sides by Sqrt 20 and Divide by Sqrt 71 (red).

Sqrt 20 (5) = | Sqrt 71 * Rate of Cl_{2 }* Sqrt 20 |

Sqrt 71 | Sqrt 20 * Sqrt 71 |

**Step 9:**

Cross out Sqrt 20 and Sqrt 71 on left

Sqrt 20 *(5) = | Sqrt 71 * Rate of Cl_{2 }* Sqrt 20 |

Sqrt 71 | Sqrt 71 * Sqrt 20 |

**Step 10:**

Simplify

Sqrt 20 * (5) = | Rate of Cl_{2 } |

Sqrt 71 |

**Step 11:**

How do I do the calculations?

Answer: ((Sqrt 20) * 5) / (Sqrt 71) = 2.65

Sqrt 20 * (5) = | 2.65 |

Sqrt 71 |

**Step 12:**

What is the complete answer?

COMPLETE ANSWER: 2.65

**VIDEO Effusion Demonstrated Example 2**: If you have an unknown gas that effuses at a rate 3 times that of Br_{2} what is the molar mass of that gas? Use this periodic table link to look up the molar masses.

**Step 1:**

What information are we given?

Answer:

Rate of Unknown = 3

Rate of Br_{2} = 1

(Look up) Molar Mass of Br_{2} = 160 g/mol

**Step 2:**

What does the question ask for?

Answer: Molar Mass of Unknown = ?

**Step 3:**

How do we set up the problem?

Answer: Start with the equation

Rate of Unkno = | Sqrt MM Br_{2} |

Rate of Br_{2} |
Sqrt MM Unkno |

**Step 4:**

What can we fill in for the equation?

Answer: The information we are given (red).

3 = | Sqrt 160 |

1 | Sqrt MM Unkno |

**Step 5:**

How do we rearrange the equation?

Answer: Multiply both sides by Sqrt MM Unknown (red).

3 * Sqrt MM Unkno = | Sqrt 160 * Sqrt MM Unkno |

1 | Sqrt MM Unknown |

**Step 6:**

Cross out Sqrt MM Unkown on right side.

3 * Sqrt MM Unkno = | Sqrt 160 * Sqrt MM Unkno |

1 | Sqrt MM Unknown |

**Step 7:**

Simplify

3 * Sqrt MM Unkno = | Sqrt 160 |

1 | 1 |

**Step 8:**

How do we rearrange the equation?

Answer: Divide both sides by 3 (red).

3 * Sqrt MM Unkno = | Sqrt 160 |

1 * 3 | 1 * 3 |

**Step 9:**

Cross out 3 on left side.

3 * Sqrt MM Unkno = | Sqrt 160 |

1 * 3 | 1 * 3 |

**Step 10:**

Simplify

Sqrt MM Unkno = | Sqrt 160 |

1 | 3 |

**Step 11:**

How do I do the calculations to help further simplify?

Answer: Sqrt 160 / 3 = 4.216

Sqrt MM Unkno = | 4.216 |

**Step 12:**

Square both sides causing the elimination of the square root on the left.

(Sqrt MM Unkno)^{2} = |
(4.216)^{2} |

**Step 13:**

Simplify

MM Unkno = | (4.216)^{2} |

**Step 14:**

How do I do the calculations?

Answer : (4.216)^{2} = 17.78

**Step 15:**

What is the complete answer?

COMPLETE ANSWER: 17.78 g/mol

**VIDEO Effusion Demonstrated Example 3**: How much faster is the rate of effusion of He gas to that of CH_{4} gas? Use this periodic table link to look up the molar masses.

**Step 1:**

What information are we given?

Answer:

(Look up) Molar Mass He = 4 g/mol

(Look up) Molar Mass CH_{4} = 16 g/mol

**Step 2:**

What does the question ask for?

Answer: Because this is a ratio we can solve for both.

Rate of CH_{4} = ?

Rate of He = ?

**Step 3:**

How do we set up the problem?

Answer: Start with the equation

Rate of He = | Sqrt MM CH_{4 } |

Rate of CH_{4} |
Sqrt MM He |

**Step 4:**

What can we fill in for the equation?

Answer: The information we are given (red).

Rate of He = | Sqrt 16 |

Rate of CH_{4} |
Sqrt 4 |

**Step 5:**

How do we simplify the math?

Answer: Apply the square roots (sqrt)

sqrt 16 becomes 4……..sqrt 4 becomes 2

Rate of He = | 4 |

Rate of CH_{4} |
2 |

**Step 6:**

How do I do the calculations?

Answer: 4 / 2 = 2

Rate of He = | 2 |

Rate of CH_{4} |
1 |

**Step 7:**

What is the complete answer?

COMPLETE ANSWER: Helium is twice the rate of CH_{4}……or 2 to 1

**PRACTICE PROBLEMS**: Solve these effusion problems. Use this periodic table link to look up the molar masses.

If the rate of effusion of N_{2} gas is 20. What is the rate of effusion of F_{2} gas?

Answer: 17.17

If you have an unknown gas that effuses at a rate 2.3 times that of Cl_{2} what is the molar mass of that gas?

Answer: 13.42

How much faster is the rate of effusion of H_{2}S gas to that of Ar gas?

Answer: 1.08 times faster

**Now try the gas laws worksheets**