EMPIRICAL FORMULAS EMPIRICAL SIMPLEST RATIO OF ATOMS IN
- Slides: 36
EMPIRICAL FORMULAS EMPIRICAL = SIMPLEST RATIO OF ATOMS IN A COMPOUND C 8 H 8 = CH COMPLEX = EMPIRICAL
Steps for empirical formulas given B= 78. 1% H= 21. 9% convert to grams find moles mole ratio formula
Steps for empirical formulas given convert to grams B= 78. 1% 78. 1 grams H= 21. 9% 21. 9 grams find moles mole ratio formula
Steps for empirical formulas given convert to grams find moles B= 78. 1% 78. 1 grams 78. 1 /10. 8 =7. 22 moles H= 21. 9% 21. 9 grams 21. 9/ 1. 01 =21. 7 moles mole ratio formula
Steps for empirical formulas given convert to grams find moles mole ratio formula B= 78. 1% 78. 1 grams 78. 1 /10. 8 7. 22 M = 1 =7. 22 M moles H= 21. 9% 21. 9 grams 21. 9/ 1. 01 21. 7 M = 3 =21. 7 7. 22 M moles
Steps for empirical formulas given convert to grams find moles mole ratio formula B= 78. 1% 78. 1 grams 78. 1 /10. 8 7. 22 M = 1 =7. 22 M moles B 1 H= 21. 9% 21. 9 grams 21. 9/ 1. 01 21. 7 M = 3 =21. 7 7. 22 M moles H 3
Steps for empirical formulas given convert to grams P = 4. 43 g O = 5. 72 g find moles mole ratio formula
Steps for empirical formulas given convert to grams find moles P = 4. 43 g 4. 43 / 31. 0 = 0. 143 m O = 5. 72 g 5. 72 / 16. 0 = 0. 357 m mole ratio formula
Steps for empirical formulas given convert to grams find moles P = 4. 43 g 4. 43 / 31. 0 = 0. 143 m O = 5. 72 g 5. 72 / 16. 0 = 0. 357 m mole ratio formula 0. 143 =1 0. 357 0. 143 =2. 5
Steps for empirical formulas given convert to grams find moles P = 4. 43 g 4. 43 / 31. 0 = 0. 143 m O = 5. 72 g 5. 72 / 16. 0 = 0. 357 m mole ratio formula 0. 143 =1 P 1 0. 357 0. 143 =2. 5 O 2. 5
Example #2 WE HAVE TO FIND A NUMBER TO MULTIPLE BOTH NUMBERS BY SO THAT THEY BECOME WHOLE NUMBERS. P = 1 AND O = 2. 5 SINCE. 5 = 1/2 THEN MULTIPLY BOTH BY 2 P = 2 AND O = 5
Steps for empirical formulas given convert to grams find moles P = 4. 43 g 4. 43 / 31. 0 = 0. 143 m O = 5. 72 g 5. 72 / 16. 0 = 0. 357 m mole ratio formula 0. 143 =1 2 X P 1 0. 357 0. 143 =2. 5 2 X O 2. 5 P 2 O 5
Example #1 STEP 4 THE WHOLE NUMBERS BECOME THE SUBSCRIPTS FOR THE FORMULA. P=2 O=5 P 2 O 5
DECIMALS = FRACTIONS. 5 = 1/2. 3 = 1/3. 667 = 2/3. 75 = 3/4. 25 = 1/4. 20 = 1/5 MULTIPLY BY 2 MULTIPLY BY 3 MULTIPLY BY 4 MULTIPLY BY 5
Steps for empirical formulas given convert to grams find moles mole ratio formula
Steps for empirical formulas given Ni = 20% Br = 80% convert to grams find moles mole ratio formula
Steps for empirical formulas given convert to grams Ni = 20% 20 g Br = 80% 80 g find moles mole ratio formula
Steps for empirical formulas given Ni = 20% convert to grams find moles 20 g 20/ 58. 7 =. 341 m mole ratio formula From periodic table Br = 80% 80 g 80/ 79. 9 = 1. 00 m
Steps for empirical formulas given Ni = 20% convert to grams 20 g find moles mole ratio formula 20/ 58. 7. 341/. 341 =. 341 m =1 From periodic table Br = 80% 80 g 80/ 79. 9 1. 00/. 341 = 1. 00 m =3
Steps for empirical formulas given Ni = 20% convert to grams 20 g find moles mole ratio formula 20/ 58. 7. 341/. 341 =. 341 m =1 Ni =1 From periodic table Br = 80% 80 g 80/ 79. 9 1. 00/. 341 = 1. 00 m =3 Br = 3 Ni Br 3
Determining Empirical Formulas: Homework 1) C H 4 2) K Cl 3) Al P O 4 4) Mg Br 2 5) Na 2 S O 4 Cu S O 4 5 H 2 O 6)
MOLECULAR FORMULAS ARE MORE COMPLEX RATIOS. SUCH AS B 3 H 9. EMPIRICAL FORMULAS ARE THE SIMPLEST RATIOS. FOR EXAMPLE BH 3
EXAMPLE #1 TWO THINGS MUST BE GIVEN IN THESE TYPES OF PROBLEMS 1) THE EMPIRICAL FORMULA = BH 3 2) THE MOLECULAR MASS = 28 grams. STEP 1 DETERMINE THE EMPIRICAL MASS FROM THE EMPIRICAL FORMULA BH 3 =13. 8 ROUND OFF TO 14
Example #1 STEP #2 DIVIDE THE MOLECULAR MASS BY THE EMPIRICAL MASS. THIS SHOULD ALWAYS GIVE YOU A WHOLE NUMBER. 28 GRAMS = 2 14 GRAMS
Example #1 STEP #3 THE WHOLE NUMBER IS MULTIPLIED BY THE SUBSCRIPTS IN THE EMPIRICAL FORMULA TO GIVE THE MOLECULAR FORMULA. EF= BH 3 X 2 = B 2 H 6=MF
EMPIRICAL FORMULA EF = MOLECULAR FORMULA MF = EMPIRICAL MASS EM = MOLECULAR MASS MM =
EMPIRICAL FORMULA EF = BH 3 EMPIRICAL MASS EM = B 1 x 10. 8 = 10. 8 H 3 X 1. 01 = 3. 03 13. 8= 14 MOLECULAR FORMULA MF = MOLECULAR MASS MM =
EMPIRICAL FORMULA EF = BH 3 EMPIRICAL MASS EM = B 1 x 10. 8 = 10. 8 H 3 X 1. 01 = 3. 03 13. 8= 14 MOLECULAR FORMULA MF = MOLECULAR MASS MM = 28 28/14 = 2
EMPIRICAL FORMULA EF = BH 3 EMPIRICAL MASS EM = B 1 x 10. 8 = 10. 8 H 3 X 1. 01 = 3. 03 13. 8= 14 MOLECULAR FORMULA MF = B 2 H 6 MOLECULAR MASS MM = 28 28/14 = 2
EMPIRICAL FORMULA EF = BH 3 EMPIRICAL MASS EM = B 1 x 10. 8 = 10. 8 H 3 X 1. 01 = 3. 03 13. 8= 14 MOLECULAR FORMULA MF = B 2 H 6 MOLECULAR MASS MM = 28 28/14 = 2
EMPIRICAL FORMULA EF = BH 3 EMPIRICAL MASS EM = B 1 x 10. 8 = 10. 8 H 3 X 1. 01 = 3. 03 13. 8= 14 MOLECULAR FORMULA MF = B 2 H 6 MOLECULAR MASS MM = 28 28/14 = 2
EMPIRICAL FORMULA EF = CH MOLECULAR FORMULA MF = EMPIRICAL MASS EM = MOLECULAR MASS MM = 78
EMPIRICAL FORMULA EF = CH MOLECULAR FORMULA EMPIRICAL MASS EM =13 MOLECULAR MASS MM = 78 MF =C 6 H 6 78 / 13 = 6
EMPIRICAL FORMULA EF = HO MOLECULAR FORMULA MF = EMPIRICAL MASS EM = MOLECULAR MASS MM = 34
EMPIRICAL FORMULA EF = HO MOLECULAR FORMULA EMPIRICAL MASS EM = 17 MOLECULAR MASS MM = 34 MF =H 2 O 2 34/17 = 2
Molecular Formulas HW 1) N 2 O 4 2) C 5 H 10 3) C 2 H 4 O 2 4) C 4 H 10 O 5) C 4 H 8 O 2
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