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Molecular Formulae

Molecular Formulae. Once the simplest formula of a compound is known, the Molar Mass can be used to determine the actual formula ( molecular formula ).

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Molecular Formulae

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  1. Molecular Formulae Once the simplest formula of a compound is known, the Molar Mass can be used to determine the actual formula (molecular formula). The simplest formula only reveals the ratio of the elements in the formula. There are an infinite number of possible molecular formulas for any given simplest formula.

  2. For instance if the simplest formula (SF) isCH3 possible molecular formulas are C2H6 C3H9 C4H12 C5H15 C6H18 C7H21 Each of these molecular formulas (MF) can be distinguished by it’s Molar Mass 30 g/mol If theSFis CH3 andtheMMis90 g/moltheMFis C6H18 45 g/mol 60 g/mol 75 g/mol 90 g/mol 105 g/mol

  3. However, if the simplest formula (SF)of CH3 has the Molar mass is 45 g/mol C2H6 C3H9 C4H12 C5H15 C6H18 C7H21 30 g/mol the MFis C3H9 45 g/mol 60 g/mol 75 g/mol 90 g/mol 105 g/mol

  4. 1 Find MF if SF is CH2 and M is 56 g/mol 2. Find MF if SF is CH2O and M is 180 g/mol 3. Find MF if SF is CCl2H4 and M is 434.7 g/mol 4. Find MF if 12.65% C, 84.16% Br, rest is H and M is 284.8 g/mol 5. Find MF if 24.49% C, 10.27% H, rest O and M is 196.2 g/mol

  5. However, if the simplest formula (SF)of CH3 has the Molar mass is 75 g/mol C2H6 C3H9 C4H12 C5H15 C6H18 C7H21 30 g/mol the MFis C5H15 45 g/mol 60 g/mol 75 g/mol 90 g/mol 105 g/mol

  6. If the simplest formula (SF) is CH2 possible molecular formulas are C2H4 C3H6 C4H8 C5H10 C6H12 C7H14 28 g/mol If the SF is CH2 and the MM is 70 g/mol the MF is 42 g/mol 56 g/mol 70 g/mol 84 g/mol C5H10 98 g/mol Is there a faster way of doing this?

  7. The simplest formula is CH2 and the molar mass of the molecular formula is 70 g/mol, Find the MF. 1. Find the MM of the SF MM of CH2 is 14 g/mol 2. Divide the MM of MF by the MM of the SF. 70 g/mol / 14 g/mol = 5 3. Multiply the SF by this multiple. 5 x CH2 = C5H10

  8. Here’s another: The simplest formula is CH and the molar mass of the MF is 78 g/mol, Find the MF. 1. Find the MM of the SF MM of CH is 13 g/mol 2. Divide the MM of MF by the MM of the SF. 78 g/mol  13 g/mol = 6 3. Multiply the SF by this multiple. 6 x CH= C6H6

  9. Find the MF of a compound which is 52.14 % carbon, 13.13 % hydrogen and oxygen and the M is 138 g/mol. What 2 pieces of information are needed before this question is solved? SF and MM To find the SF masses of each element are needed. In 100 g of compound this compound there are 52.14 g of C, 13.13 g of H, 34.73 g O. Find nC, nH,nO

  10. Find the MF of a compound which is 52.14 % carbon, 13.13% hydrogen and oxygen. The M is 138 g/mol. nC =52.14 g / 12 g/mol = 4.35 mol C nH =13.13 g / 1 g/mol = 13.13 mol H nO =34.73 g / 16 g/mol = 2.17 mol O Compare the molar quantities to O. 4.35 mol C/2.17 mol O = 2 mol C/mol O 13.13 mol H/2.17 mol O = 6 mol H/mol O Simplest Formula (SF)is C2H6O

  11. Find the MF of a compound which is 52.14 % carbon, 13.13% hydrogen and oxygen. The M is 138 g/mol. Simplest Formula (SF)is C2H6O MM of SF is 46 g/mol 138 g/mol/46 g/mol is 3. 3 x C2H6O = C6H18O3

  12. Find the MF of a compound which is 52.14 % carbon, 13.13% hydrogen and oxygen if M is 92 g/mol. SFis C2H6O MM of SF is 46 g/mol MM of MF / MM of SF is 92 g/mol / 46 g/mol 2 2 x C2H6O = C4H12O2

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