Upload
doandung
View
232
Download
3
Embed Size (px)
Citation preview
Chemistry: Empirical Formula Guided Inquiry v2
Empirical Formula vs. Molecular Formula: What’s the difference?
Molecular Formula – shows us the number and type of atoms that make up a molecule.
Empirical Formula -‐ shows the simplest ratio of elements in a molecular compound.
Sometimes the empirical formula and molecular formula are the same. Other times they are different. Let’s look at the molecular and empirical formulae for water and hydrogen peroxide:
Water Hydrogen Peroxide
Molecular Formula H2O H2O2
Empirical Formula H2O HO
For water, the molecule is made up of two hydrogen atoms and one oxygen atom, so its molecular formula is H2O. This also represents the simplest ratio of atoms in the molecule, so its empirical formula is H2O. So water’s molecular and empirical formulae are the same. However, they are different for hydrogen peroxide. A hydrogen peroxide molecule is made of two hydrogen atoms and two oxygen atoms, so its molecular formula is H2O2. The ratio of hydrogen atoms to oxygen atoms is 1:1, so hydrogen peroxide’s empirical formula is HO.
1. Explain why the molecular formula for hydrogen peroxide is H2O2.
2. Explain why the empirical formula for hydrogen peroxide is HO.
3. Write the molecular and empirical formulas for each of the compounds below.
Ethene Propene
Molecular Formula
Empirical Formula
4. Why do these two compounds have different molecular formulas, but the same empirical formula?
5. Write the molecular and empirical formulas for each of the compounds below.
Ethane Propane
Molecular Formula
Empirical Formula
6. Do these two compounds have the same or different empirical formulae? Explain why.
7. Write the molecular and empirical formulas for each of the compounds below.
Formaldehyde Acetic Acid Glucose
Molecular Formula
Empirical Formula
8. Do these three compounds have the same or different empirical formulae? Explain why.
9. Explain how you can determine a compound’s empirical formula from its molecular formula.
STOP – Take the online Empirical Quick Quiz #1. When you are finished see your teacher for your quiz score. Bring your guided inquiry with you.
Empirical Formulae: Why do we need to know? When analytical chemists analyze an unknown substance, they don’t know its molecular formula. But they can use their laboratory equipment to determine the elements and how much of each one are in a compound. Determining the empirical formula of an unknown compound is the first step to determining its molecular formula and identity.
How to Determine Empirical Formulae The subscripts in an empirical formula represent the ratio of atoms in a compound. For example, ethene’s empirical formula of CH2 means that for every carbon atom there are two hydrogen atoms. If we know the number of moles of each atom in a compound we can determine the empirical formula.
Example #1: A sample of an unknown compound was sent to an analytical lab to determine its composition. The lab analysis shows that the sample’s percent composition by mass is 80.0% carbon and 20.0% hydrogen. What is the compound’s empirical formula?
Step 1 – Convert the percentages of each element to grams by assuming we have a 100.0 gram sample:
𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶 = 90.0%×100.0 𝑔 = 80.0 𝑔 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻 = 10.0%×100.0 𝑔 = 10.0 𝑔
Step 2 – Convert the grams of each element to moles using the molar mass of each element:
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶 = 90.0 𝑔× ! !"#!".! !
= 7.50 𝑚𝑜𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻 = 10.0 𝑔× ! !"#!.! !
= 10.0 𝑚𝑜𝑙
Step 3 – Convert the number of moles of each element to a whole number ratio by dividing the smallest number into the other numbers:
𝐻 𝑡𝑜 𝐶 𝑟𝑎𝑡𝑖𝑜 = !".! !"#!.!" !"#
= 1.33 C 𝑡𝑜 𝐶 𝑟𝑎𝑡𝑖𝑜 = !.!" !"#!.!" !"!
= 1.00
1.33 is not a whole number so we need to multiply each ratio by 3 to get whole numbers
ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 𝑟𝑎𝑡𝑖𝑜 = 1.33×3 = 4 𝑐𝑎𝑟𝑏𝑜𝑛 𝑟𝑎𝑡𝑖𝑜 = 1.00×3 = 3
Step 4 – Write the empirical formula using the whole number ratios as the subscripts:
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝐶!𝐻!
Example #2: What is the empirical formula of a compound that is 75.7% tin (Sn) and 24.3% fluorine (F)? To solve this problem let’s assume we have a 100.0 gram sample of the compound.
10. What is the mass of tin in the 100.0 g sample?
11. What is the mass of fluorine in the 100.0 g sample?
12. How many moles of tin are in this sample?
13. How many moles of fluorine are in this sample?
14. What is the simplest whole number ratio of the moles of the elements?
15. Write the empirical formula using the numbers from the mole ratio for the subscripts. Remember that the metal symbol comes first followed by the non-‐metal symbol; and we don’t write the subscript “1”.
Example #3: What is the empirical formula of a compound that is 70.0% iron (Fe) and 30.0% oxygen (O)? Let’s assume we have a 100.0 gram sample of the compound.
16. What is the mass of iron in the 100.0 g sample?
17. What is the mass of oxygen in the 100.0 g sample?
18. How many moles of iron are in this sample?
19. How many moles of oxygen are in this sample?
20. What is the simplest whole number ratio of the moles of the elements?
21. Write the empirical formula using the numbers from the mole ratio for the subscripts.
22. Summarize the 4-‐step process for determining the empirical formula from the percent composition by mass.
STOP – Take the online Empirical Quick Quiz #2. When you are finished see your teacher for your quiz score. Bring your guided inquiry with you.
Determining Molecular Formulas If we know the molar mass of a compound we can determine its molecular formula from the empirical formula. Remember, the empirical formula give us the simplest ratio of atoms in a compound. The molecular formula gives us the actual number of atoms in in the molecule.
Example #4 – The analytical lab report shows that an unknown compound has an empirical formula of CH2O and a molar mass of 180.0 grams. What is the compound’s molecular formula?
Step 1 – Determine the molar mass of the empirical formula
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 12.0𝑔×1 + 1.0𝑔×2 + 16.0𝑔×1 = 30.0 𝑔
Step 2 – Divide the empirical molar mass into the molar mass to get a whole number ratio:
180.0 𝑔 ÷ 30.0 𝑔 = 6
Step 3 – Determine the molecular formula by multiplying the subscripts in the empirical formula by the whole number ratio:
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝐶!×!𝐻!×!𝑂!×! = 𝐶!𝐻!"𝑂!
Example #5 – What is the molecular formula of a compound that has an empirical formula of NS and a molar mass of 184 grams.
23. What is the mass of one mole of the empirical formula?
24. Is the empirical formula mass larger or smaller than the molar mass of 184 grams?
25. How many times heavier is the molar mass than the empirical formula mass?
26. Use this multiplier to adjust the subscripts of the empirical formula so its mass is equal to the molar mass. This is the molecular formula.
Example #6 – What is the molecular formula of a compound that has an empirical formula of CH3 and a molar mass of 30.0 grams.
27. What is the mass of one mole of the empirical formula?
28. Is the empirical formula mass larger or smaller than the molar mass of 30.0 grams?
29. How many times heavier is the molar mass than the empirical formula mass?
30. Use this multiplier to adjust the subscripts of the empirical formula so its mass is equal to the molar mass. This is the molecular formula.
31. Summarize the 3-‐step process for determining the molecular formula from the compound’s molar mass and empirical formula.
STOP – Take the online Empirical Quick Quiz #3. When you are finished see your teacher for your quiz score. Bring your guided inquiry with you.