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**Calculating Empirical Formulas**

Lowest whole-number ratio of the atoms of the elements in a compound C6H12O6 (glucose) The ratio that glucose normally has for carbon:hydrogen:oxygen is 6:12:6. The lowest ratio that glucose has for carbon:hydrogen:oxygen is 1:2:1 (each number can be divided by the smallest number in the ratio which is 6).

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**Next - Calculating empirical formulas**

The empirical formula for glucose is CH2O since this is the lowest whole-number ratio of atoms for that compound. May or may not be the same as the normal molecular formula of a compound Next - Calculating empirical formulas

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**What is the empirical formula of a compound that is 25**

What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? If 25.9% of the compound is nitrogen and 74.1% of the compound is oxygen, then a compound with a mass of 100 g has 25.9 g of nitrogen and 74.1 g of oxygen. To calculate the empirical formula, we need to relate the moles of each atom in the compound, so we need to convert the masses of the elements to moles.

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**This would mean that the ratio of nitrogen to oxygen**

is N1.85O4.63. We can divide each number in the ratio of N1.85O4.63 by 1.85 to get N1O2.50. Since we cannot have 2.50 atoms of oxygen, we must multiply through each number by 2 to even it out, getting N2O5 as our empirical formula.

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In calculating empirical formulas, remember that the number of atoms is a whole number. If the number of atoms for an element is close to a whole number (i.e., 2.1 or 2.2 or 2.8, or 2.9), you can usually round up or down to get a whole number. If you should get a number of atoms closer to 2.33 or 2.5, multiply each number in the formula by a number that gets that to a whole number. For example, if you calculated 2.33, you would multiply this by 3 to get a value of 7 for that number.

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Give it a try Determine the empirical formula for a compound containing 7.8% carbon and 92.2% chlorine. 7.8 g C 1 mol C 12.0 g C = 0.65 mol C 92.2 g Cl 1mol Cl 35.5 g Cl = mol Cl C0.65Cl2.60 divide each by 0.65 Final answer = CCl4

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**Empirical vs. Molecular Formula**

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**Calculating Molecular Formulas**

Although sometimes a molecular formula may be the same as a molecule’s empirical formula, like in carbon dioxide (CO2), we have seen that the empirical formula for glucose is not the same as its molecular formula. One can determine the molecular formula of a compound by knowing its empirical formula and its mass. Next - Example

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Calculate the molecular formula of the compound whose molar mass is g and empirical formula is CH2O. We know that the molecular formula will have a molar mass of g. We also know, by calculating the gmm of CH2O, that CH2O has an empirical formula mass (efm) = g CH2O. Now, in order to figure out what we must multiply each number in the empirical formula by, we must figure out by what number we must multiply the empirical formula mass to get the molecular formula mass. To get from to ,

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Therefore, we must multiply each number of atoms in CH2O by 6 to get the molecular formula of C6H12O6. You can double-check your answer by recalculating the molar mass of C6H12O6. gmm C6H12O6 = 6 x g + 12 x g + 6 x g = g C6H12O6 This agrees with the molar mass we were given, so the molecular formula we calculated is correct.

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Give it a try Determine the molecular formula of a compound that is 40.0% C, 6.6% H, and 53.4% O and the molar mass is 120.0g. mol C = mol H = 6.6 mol O = 3.34 divide each by 3.33 C1H2O1 Empirical Formula weight is 30.0 Take divided by 30.0 = 4 Than multiply your subscripts. Final Answer is C4H8O4

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Example (toughy) 1.00 g of menthol on combustion yields g of H2O and g of CO2. What is the empirical formula? Solution: 1.161 g H2O g H = mol H 2.818 g CO2 g C = mol C difference = g O = mol O ratios: H/O = 20.4 and C/O =10.1 Therefore, C10H20O

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Empirical and Molecular Formulas. Formaldehyde CH 2 O Acetic acid C 2 H 4 O 2 Gylceradehyde C 3 H 6 O 3 40% C; 6.7% H; 53.3% O.

Empirical and Molecular Formulas. Formaldehyde CH 2 O Acetic acid C 2 H 4 O 2 Gylceradehyde C 3 H 6 O 3 40% C; 6.7% H; 53.3% O.

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