Empirical and Molecular Formulas Empirical vs Molecular Formula
![Empirical and Molecular Formulas Empirical and Molecular Formulas](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-1.jpg)
![Empirical vs Molecular Formula • The Molecular Formula (MF) gives the actual number of Empirical vs Molecular Formula • The Molecular Formula (MF) gives the actual number of](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-2.jpg)
![Types of Formulas The formulas for compounds can be expressed as an empirical formula Types of Formulas The formulas for compounds can be expressed as an empirical formula](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-3.jpg)
![• An empirical formula represents the simplest whole number ratio of the atoms • An empirical formula represents the simplest whole number ratio of the atoms](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-4.jpg)
![EF vs. MF When you have an ionic compound: the EF = MF For EF vs. MF When you have an ionic compound: the EF = MF For](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-5.jpg)
![Determining the EF Remember: The EF is the lowest whole-number ratio of the moles Determining the EF Remember: The EF is the lowest whole-number ratio of the moles](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-6.jpg)
![Determining the EF If a compound consists of: 62. 1% C 13. 8% H Determining the EF If a compound consists of: 62. 1% C 13. 8% H](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-7.jpg)
![Determining the EF In order to determine the ratio of C: H: N, we Determining the EF In order to determine the ratio of C: H: N, we](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-8.jpg)
![Determining the EF Now that you know how many grams of each atom you Determining the EF Now that you know how many grams of each atom you](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-9.jpg)
![Determining the EF Now that you have the mol ratio, you need to make Determining the EF Now that you have the mol ratio, you need to make](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-10.jpg)
![Determining the EF This results in 3 mol C, 8 mol H and 1 Determining the EF This results in 3 mol C, 8 mol H and 1](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-11.jpg)
![Learning Check EF-1 A. What is the empirical formula for C 4 H 8? Learning Check EF-1 A. What is the empirical formula for C 4 H 8?](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-12.jpg)
![Solution EF-1 A. What is the empirical formula for C 4 H 8? 2) Solution EF-1 A. What is the empirical formula for C 4 H 8? 2)](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-13.jpg)
![Learning Check EF-2 If the molecular formula has 4 atoms of N, what is Learning Check EF-2 If the molecular formula has 4 atoms of N, what is](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-14.jpg)
![Solution EF-2 If the molecular formula has 4 atoms of N, what is the Solution EF-2 If the molecular formula has 4 atoms of N, what is the](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-15.jpg)
![Empirical and Molecular Formulas molar mass = simplest mass a whole number = n Empirical and Molecular Formulas molar mass = simplest mass a whole number = n](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-16.jpg)
![EF to MF To go from the EF to the MF, you need two EF to MF To go from the EF to the MF, you need two](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-17.jpg)
![EF to MF Example: You found the EF to be HO, and the MFmass EF to MF Example: You found the EF to be HO, and the MFmass](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-18.jpg)
![Learning Check EF-3 A compound has a formula mass of 176. 0 and an Learning Check EF-3 A compound has a formula mass of 176. 0 and an](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-19.jpg)
![Solution EF-3 A compound has a formula mass of 176. 0 and an empirical Solution EF-3 A compound has a formula mass of 176. 0 and an empirical](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-20.jpg)
![Learning Check EF-4 If there are 192. 0 g of O in the molecular Learning Check EF-4 If there are 192. 0 g of O in the molecular](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-21.jpg)
![Solution EF-4 If there are 192. 0 g of O in the molecular formula, Solution EF-4 If there are 192. 0 g of O in the molecular formula,](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-22.jpg)
![Finding the Molecular Formula A compound is Cl 71. 65%, C 24. 27%, and Finding the Molecular Formula A compound is Cl 71. 65%, C 24. 27%, and](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-23.jpg)
![2. Calculate the number of moles of each element. 71. 65 g Cl x 2. Calculate the number of moles of each element. 71. 65 g Cl x](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-24.jpg)
![Why moles? Why do you need the number of moles of each element in Why moles? Why do you need the number of moles of each element in](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-25.jpg)
![3. Find the smallest whole number ratio by dividing each mole value by the 3. Find the smallest whole number ratio by dividing each mole value by the](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-26.jpg)
![5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49. 5 6. 5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49. 5 6.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-27.jpg)
![Learning Check EF-5 Aspirin is 60. 0% C, 4. 5 % H and 35. Learning Check EF-5 Aspirin is 60. 0% C, 4. 5 % H and 35.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-28.jpg)
![Solution EF-5 60. 0 g C x 4. 5 g H ______= ______ mol Solution EF-5 60. 0 g C x 4. 5 g H ______= ______ mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-29.jpg)
![Solution EF-5 60. 0 g C x 1 mol C = 5. 00 mol Solution EF-5 60. 0 g C x 1 mol C = 5. 00 mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-30.jpg)
![Divide by the smallest # of moles. 5. 00 mol C = ________ mol Divide by the smallest # of moles. 5. 00 mol C = ________ mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-31.jpg)
![Divide by the smallest # of moles. 5. 00 mol C = ___2. 25__ Divide by the smallest # of moles. 5. 00 mol C = ___2. 25__](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-32.jpg)
![Finding Subscripts A fraction between 0. 1 and 0. 9 must not be rounded. Finding Subscripts A fraction between 0. 1 and 0. 9 must not be rounded.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-33.jpg)
![Multiply everything x 4 C: 2. 25 mol C x 4 = 9 mol Multiply everything x 4 C: 2. 25 mol C x 4 = 9 mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-34.jpg)
![Learning Check EF-6 A compound is 27. 4% S, 12. 0% N and 60. Learning Check EF-6 A compound is 27. 4% S, 12. 0% N and 60.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-35.jpg)
![Solution EF 6 0. 853 mol S /0. 853 = 1 S 0. 857 Solution EF 6 0. 853 mol S /0. 853 = 1 S 0. 857](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-36.jpg)
- Slides: 36
![Empirical and Molecular Formulas Empirical and Molecular Formulas](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-1.jpg)
Empirical and Molecular Formulas
![Empirical vs Molecular Formula The Molecular Formula MF gives the actual number of Empirical vs Molecular Formula • The Molecular Formula (MF) gives the actual number of](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-2.jpg)
Empirical vs Molecular Formula • The Molecular Formula (MF) gives the actual number of each type of atom present. • The Empirical Formula (EF) gives the lowest wholenumber ratio of the atoms present. • Example: C 2 H 6 and C 3 H 9 – they have the same EF CH 3 yet have very different MF
![Types of Formulas The formulas for compounds can be expressed as an empirical formula Types of Formulas The formulas for compounds can be expressed as an empirical formula](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-3.jpg)
Types of Formulas The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula. Empirical Molecular (true) Name CH C 2 H 2 acetylene CH C 6 H 6 benzene CO 2 CH 2 O C 5 H 10 O 5 Timberlake Lecture. PLUS carbon dioxide ribose 3
![An empirical formula represents the simplest whole number ratio of the atoms • An empirical formula represents the simplest whole number ratio of the atoms](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-4.jpg)
• An empirical formula represents the simplest whole number ratio of the atoms in a compound. • The molecular formula is the true or actual ratio of the atoms in a compound. Timberlake Lecture. PLUS 4
![EF vs MF When you have an ionic compound the EF MF For EF vs. MF When you have an ionic compound: the EF = MF For](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-5.jpg)
EF vs. MF When you have an ionic compound: the EF = MF For some molecular compounds: the EF = MF, but not always
![Determining the EF Remember The EF is the lowest wholenumber ratio of the moles Determining the EF Remember: The EF is the lowest whole-number ratio of the moles](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-6.jpg)
Determining the EF Remember: The EF is the lowest whole-number ratio of the moles of each atom present. Example: CH 4 has 1 mol C atoms 4 mol H atoms
![Determining the EF If a compound consists of 62 1 C 13 8 H Determining the EF If a compound consists of: 62. 1% C 13. 8% H](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-7.jpg)
Determining the EF If a compound consists of: 62. 1% C 13. 8% H 24. 1% N The percentages are based on MASS not MOLES We can compare moles, not masses
![Determining the EF In order to determine the ratio of C H N we Determining the EF In order to determine the ratio of C: H: N, we](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-8.jpg)
Determining the EF In order to determine the ratio of C: H: N, we need to know the mole ratio Step 1: Convert % of each into grams Make it easy on yourself, assume a sample size of 100. 00 g 62. 1 g C 13. 8 g H 24. 1 g N
![Determining the EF Now that you know how many grams of each atom you Determining the EF Now that you know how many grams of each atom you](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-9.jpg)
Determining the EF Now that you know how many grams of each atom you have: Step 2: Convert grams to moles using the molar mass of each 62. 1 g C 13. 8 g H 24. 1 g N
![Determining the EF Now that you have the mol ratio you need to make Determining the EF Now that you have the mol ratio, you need to make](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-10.jpg)
Determining the EF Now that you have the mol ratio, you need to make them Whole-Numbers. Step 3: Divide each mol by the smallest mol value from step #2 5. 17 mol C 13. 7 mol H 1. 72 mol N
![Determining the EF This results in 3 mol C 8 mol H and 1 Determining the EF This results in 3 mol C, 8 mol H and 1](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-11.jpg)
Determining the EF This results in 3 mol C, 8 mol H and 1 mol N therefore the EF = C 3 H 8 N
![Learning Check EF1 A What is the empirical formula for C 4 H 8 Learning Check EF-1 A. What is the empirical formula for C 4 H 8?](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-12.jpg)
Learning Check EF-1 A. What is the empirical formula for C 4 H 8? 1) C 2 H 4 2) CH 2 3) CH B. What is the empirical formula for C 8 H 14? 1) C 4 H 7 2) C 6 H 12 3) C 8 H 14 C. What is a molecular formula for CH 2 O? 1) CH 2 O 2) C 2 H 4 O 2 Timberlake Lecture. PLUS 3) C 3 H 6 O 3 12
![Solution EF1 A What is the empirical formula for C 4 H 8 2 Solution EF-1 A. What is the empirical formula for C 4 H 8? 2)](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-13.jpg)
Solution EF-1 A. What is the empirical formula for C 4 H 8? 2) CH 2 B. What is the empirical formula for C 8 H 14? 1) C 4 H 7 C. What is a molecular formula for CH 2 O? 1) CH 2 O 2) C 2 H 4 O 2 Timberlake Lecture. PLUS 3) C 3 H 6 O 3 13
![Learning Check EF2 If the molecular formula has 4 atoms of N what is Learning Check EF-2 If the molecular formula has 4 atoms of N, what is](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-14.jpg)
Learning Check EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 1) SN 2) SN 4 3) S 4 N 4 Timberlake Lecture. PLUS 14
![Solution EF2 If the molecular formula has 4 atoms of N what is the Solution EF-2 If the molecular formula has 4 atoms of N, what is the](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-15.jpg)
Solution EF-2 If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain. 3) S 4 N 4 If the actual formula has 4 atoms of N, and S is related 1: 1, then there must also be 4 atoms of S. Timberlake Lecture. PLUS 15
![Empirical and Molecular Formulas molar mass simplest mass a whole number n Empirical and Molecular Formulas molar mass = simplest mass a whole number = n](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-16.jpg)
Empirical and Molecular Formulas molar mass = simplest mass a whole number = n n = 1 molar mass = empirical mass molecular formula = empirical formula n = 2 molar mass = 2 x empirical mass molecular formula = 2 x empirical formula molecular formula = or > empirical formula 16 Timberlake Lecture. PLUS
![EF to MF To go from the EF to the MF you need two EF to MF To go from the EF to the MF, you need two](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-17.jpg)
EF to MF To go from the EF to the MF, you need two additional pieces of information: 1 – calculate the mass from your EF 2 – You must be given the mass of the MF (X) EFmass = MFmass X = MFmass / EFmass
![EF to MF Example You found the EF to be HO and the MFmass EF to MF Example: You found the EF to be HO, and the MFmass](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-18.jpg)
EF to MF Example: You found the EF to be HO, and the MFmass = 34. 02 g 1. 2. 3. 4. Calculate the EFmass = 17. 01 g Calculate X = 34. 02 g / 17. 01 g X=2 EF = HO MF = H 2 O 2
![Learning Check EF3 A compound has a formula mass of 176 0 and an Learning Check EF-3 A compound has a formula mass of 176. 0 and an](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-19.jpg)
Learning Check EF-3 A compound has a formula mass of 176. 0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 1) C 3 H 4 O 3 2) C 6 H 8 O 6 3) C 9 H 12 O 9 Timberlake Lecture. PLUS 19
![Solution EF3 A compound has a formula mass of 176 0 and an empirical Solution EF-3 A compound has a formula mass of 176. 0 and an empirical](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-20.jpg)
Solution EF-3 A compound has a formula mass of 176. 0 and an empirical formula of C 3 H 4 O 3. What is the molecular formula? 2) C 6 H 8 O 6 C 3 H 4 O 3 = 88. 0 g/EF 176. 0 g = 2. 00 88. 0 Timberlake Lecture. PLUS 20
![Learning Check EF4 If there are 192 0 g of O in the molecular Learning Check EF-4 If there are 192. 0 g of O in the molecular](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-21.jpg)
Learning Check EF-4 If there are 192. 0 g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4? 1) C 7 H 6 O 4 2) C 14 H 12 O 8 3) C 21 H 18 O 12 Timberlake Lecture. PLUS 21
![Solution EF4 If there are 192 0 g of O in the molecular formula Solution EF-4 If there are 192. 0 g of O in the molecular formula,](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-22.jpg)
Solution EF-4 If there are 192. 0 g of O in the molecular formula, what is the true formula if the EF is C 7 H 6 O 4? 3) C 21 H 18 O 12 192 g O = 3 x O 4 or 3 x C 7 H 6 O 4 Timberlake Lecture. PLUS 64. 0 g O in EF 22
![Finding the Molecular Formula A compound is Cl 71 65 C 24 27 and Finding the Molecular Formula A compound is Cl 71. 65%, C 24. 27%, and](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-23.jpg)
Finding the Molecular Formula A compound is Cl 71. 65%, C 24. 27%, and H 4. 07%. What are the empirical and molecular formulas? The molar mass is known to be 99. 0 g/mol. 1. State mass percents as grams in a 100. 00 -g sample of the compound. Cl 71. 65 g C 24. 27 g H 4. 07 g Timberlake Lecture. PLUS 23
![2 Calculate the number of moles of each element 71 65 g Cl x 2. Calculate the number of moles of each element. 71. 65 g Cl x](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-24.jpg)
2. Calculate the number of moles of each element. 71. 65 g Cl x 1 mol Cl = 2. 02 mol Cl 35. 5 g Cl 24. 27 g C x 1 mol C 12. 0 g C = 2. 02 mol C 4. 07 g H x 1 mol H 1. 01 g H = Timberlake Lecture. PLUS 4. 04 mol H 24
![Why moles Why do you need the number of moles of each element in Why moles? Why do you need the number of moles of each element in](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-25.jpg)
Why moles? Why do you need the number of moles of each element in the compound? Timberlake Lecture. PLUS 25
![3 Find the smallest whole number ratio by dividing each mole value by the 3. Find the smallest whole number ratio by dividing each mole value by the](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-26.jpg)
3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values: Cl: 2. 02 = 1 Cl 2. 02 C: H: 2. 02 = 1 C 4. 04 = 2 H 2. 02 4. Write the simplest or empirical formula Timberlake Lecture. PLUS 26 CH 2 Cl
![5 EM empirical mass 1C 2H 1Cl 49 5 6 5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49. 5 6.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-27.jpg)
5. EM (empirical mass) = 1(C) + 2(H) + 1(Cl) = 49. 5 6. n = molar mass/empirical mass Molar mass EM = 99. 0 g/mol 49. 5 g/EM = n=2 7. Molecular formula (CH 2 Cl)2 = C 2 H 4 Cl 2 Timberlake Lecture. PLUS 27
![Learning Check EF5 Aspirin is 60 0 C 4 5 H and 35 Learning Check EF-5 Aspirin is 60. 0% C, 4. 5 % H and 35.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-28.jpg)
Learning Check EF-5 Aspirin is 60. 0% C, 4. 5 % H and 35. 5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60. 0 g C, 4. 5 g H, and 35. 5 g O. Timberlake Lecture. PLUS 28
![Solution EF5 60 0 g C x 4 5 g H mol Solution EF-5 60. 0 g C x 4. 5 g H ______= ______ mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-29.jpg)
Solution EF-5 60. 0 g C x 4. 5 g H ______= ______ mol C x ______ = _______mol H 35. 5 g O x ______ = _______mol O Timberlake Lecture. PLUS 29
![Solution EF5 60 0 g C x 1 mol C 5 00 mol Solution EF-5 60. 0 g C x 1 mol C = 5. 00 mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-30.jpg)
Solution EF-5 60. 0 g C x 1 mol C = 5. 00 mol C = 4. 5 mol H = 2. 22 mol O 12. 0 g C 4. 5 g H x 1 mol H 1. 01 g H 35. 5 g O x 1 mol O 16. 0 g O Timberlake Lecture. PLUS 30
![Divide by the smallest of moles 5 00 mol C mol Divide by the smallest # of moles. 5. 00 mol C = ________ mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-31.jpg)
Divide by the smallest # of moles. 5. 00 mol C = ________ mol O 4. 5 mol H = ______ mol O ________ 2. 22 mol O = ________ mol O Are are the results whole numbers? _____ Timberlake Lecture. PLUS 31
![Divide by the smallest of moles 5 00 mol C 2 25 Divide by the smallest # of moles. 5. 00 mol C = ___2. 25__](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-32.jpg)
Divide by the smallest # of moles. 5. 00 mol C = ___2. 25__ 2. 22 mol O 4. 5 mol H 2. 22 mol O = ___2. 00__ 2. 22 mol O = ___1. 00__ 2. 22 mol O Are are the results whole numbers? _____ Timberlake Lecture. PLUS 32
![Finding Subscripts A fraction between 0 1 and 0 9 must not be rounded Finding Subscripts A fraction between 0. 1 and 0. 9 must not be rounded.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-33.jpg)
Finding Subscripts A fraction between 0. 1 and 0. 9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts. (1/2) (1/3) (1/4) (3/4) 0. 5 0. 333 0. 25 0. 75 x 2 x 3 x 4 = = Timberlake Lecture. PLUS 1 1 1 3 33
![Multiply everything x 4 C 2 25 mol C x 4 9 mol Multiply everything x 4 C: 2. 25 mol C x 4 = 9 mol](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-34.jpg)
Multiply everything x 4 C: 2. 25 mol C x 4 = 9 mol C H: 2. 0 mol H x 4 = 8 mol H O: 1. 00 mol O x 4 = 4 mol O Use the whole numbers of mols as the subscripts in the simplest formula C 9 H 8 O 4 Timberlake Lecture. PLUS 34
![Learning Check EF6 A compound is 27 4 S 12 0 N and 60 Learning Check EF-6 A compound is 27. 4% S, 12. 0% N and 60.](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-35.jpg)
Learning Check EF-6 A compound is 27. 4% S, 12. 0% N and 60. 6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula? Timberlake Lecture. PLUS 35
![Solution EF 6 0 853 mol S 0 853 1 S 0 857 Solution EF 6 0. 853 mol S /0. 853 = 1 S 0. 857](https://slidetodoc.com/presentation_image_h/752d05219f10b6497d8d9266f3452cac/image-36.jpg)
Solution EF 6 0. 853 mol S /0. 853 = 1 S 0. 857 mol N /0. 853 = 1 N 1. 71 mol Cl /0. 853 = 2 Cl Empirical formula = SNCl 2 = 117. 1 g/EF Mol. Mass/ Empirical mass 351/117. 1 = 3 Molecular formula = S 3 N 3 Cl 6 Timberlake Lecture. PLUS 36
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