Bellwork: Get out your old textbook and your ID. Get out a sheet of paper to take some notes on. Solve the following problem. Methyl alcohol, CH 3 OH, is a clean-burning, easily handled fuel. It can be made by the direct reaction of CO and H 2 (obtained from coal and water). CO(g) + 2 H 2 (g) CH 3 OH(l) Assume you start with 12.0g of H 2 and 74.5g of CO, what mass (in grams) of the excess reagent is left after the reaction is complete?
Books Back
* 1.Find the molar (or formula) mass of the compound. 2.Divide the mass from each element in the compound by the molar/formula mass. 3.Multiply the resulting decimals by 100%.
* Calculate the percent composition of caffeine, C 8 H 10 N 4 O 2.
* Calculate the percent composition of caffeine, C 8 H 10 N 4 O 2.
The empirical formula is the lowest whole-number ratio of atoms in a compound. For ionic compounds this is often the ionic formula unit as well. It is very useful in compound identification and is a part of many analysis techniques used in analytical laboratories. It can be found in several ways (just to name a few): (1) from percent composition, (2) from combustion analysis, (3) from the molecular formula, and (4) from raw mass data. It is related to the molecular formula by an integer multiplier.
Step 1 Assume 100g of sample and convert all percentages to masses (in grams). Step 2 Convert all masses to moles. Step 3 Normalize all mole values. Step 4 Use these values as subscripts in the formula. Multiply by whole numbers to get integers for all values if needed. Example Determine the empirical formula of the compound with the following percent composition: 10.4% carbon, 27.8% sulfur, and 61.7% chlorine.
Example Determine the empirical formula of the compound with the following percent composition: 10.4% carbon, 27.8% sulfur, and 61.7% chlorine.
Step 1 Assume all carbon in the sample is converted to CO 2 and that all hydrogen is converted to H 2 O. Oxygen in the sample will be split between both and mixed with outside O 2 and its amount must be calculated last. Step 2 Convert the mass of CO 2 to moles of carbon and the mass of water to moles of hydrogen. If oxygen is present, find the mass of oxygen and convert it to moles of oxygen. Step 3 Normalize all mole values. Step 4 Use these values as subscripts in the formula. Multiply by whole numbers to get integers for all values if needed. Note If other elements are present, such as sulfur or nitrogen, they are converted to their most common oxides and will appear in the mass data as such. Nitrogen will appear as NO 2 and sulfur will appear as SO 2.
Simple example: Butane is a hydrocarbon (containing only C and H) compound commonly used in lighters. Combustion of 10.0g of butane resulted in the formation of 30.3g of CO 2 and 15.5g of H 2 O. Determine the empirical formula of butane.
Simple example: Butane is a hydrocarbon (containing only C and H) compound commonly used in lighters. Combustion of 10.0g of butane resulted in the formation of 30.3g of CO 2 and 15.5g of H 2 O. Determine the empirical formula of butane.
Harder example: Ethyl butyrate, which is responsible for the characteristic smell of pineapple, is comprised of carbon, hydrogen, and oxygen. Combustion of 2.78g of ethyl butyrate produces 6.32g of CO 2 and 2.58g of H 2 O. Find the empirical formula of ethyl butyrate.
Harder example: Ethyl butyrate, which is responsible for the characteristic smell of pineapple, is comprised of carbon, hydrogen, and oxygen. Combustion of 2.78g of ethyl butyrate produces 6.32g of CO 2 and 2.58g of H 2 O. Find the empirical formula of ethyl butyrate.
Step 1 Write the correct molecular formula for the compound. Step 2 Reduce all subscripts to the lowest possible whole numbers. Example Determine the empirical formula of glucose, C 6 H 12 O 6.
Step 1 Write the correct molecular formula for the compound. Step 2 Reduce all subscripts to the lowest possible whole numbers. Example Determine the empirical formula of glucose, C 6 H 12 O 6.
Step 1 Obtain the molecular mass of the compound by some means. Step 2 Calculate the formula mass of the empirical formula. Step 3 Divide the molecular mass by the empirical formula mass. Step 4 Multiply all of the subscripts in the empirical formula by the resulting integer from step 3 and write the molecular formula with the new subscripts. Example The molecular mass of a certain diamine compound is found from TOF mass spectroscopy to be 88g/mol. The empirical formula is determined via combustion analysis to be C 2 H 6 N. What is the molecular formula of the diamine?
Example The molecular mass of a certain diamine compound is found from TOF mass spectroscopy to be 88g/mol. The empirical formula is determined via combustion analysis to be C 2 H 6 N. What is the molecular formula of the diamine?
p. 84 new textbook http://www.bozemanscience.com/ap-chem-009-massspectrometry/
Practice on sheet
Isotope name Isotope mass ( amu) Relative Abundance Silicon-28 27.98 92.21 Silicon-29 28.98 4.70 Silicon-30 29.97 3.09
Isotope name Isotope mass ( amu) Relative Abundance Silicon-28 27.98 92.21 Silicon-29 28.98 4.70 Silicon-30 29.97 3.09
Practice