Chemistry textbook chapter 3 PDF

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Chemistry textbook chapter 3; Formula mass and the mole concept for compounds; -

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For any compound the formula mass is the sum of the atomic masses of all the atoms in its chemical formula Formula mass formula; o Formula mass= (number of atoms of 1st element in the chemical formula X the atomic mass of the first element) + (number of atoms of 2nd element in the chemical formula X atomic mass of the second element) o Example; CO2 (2 is a subscript btw)  12.01 amu = C  16.00 amu= O  12.01 + 2(16.00)= 44.01 Molar mass of a compound; We know that molar mass is the mass in grams of one mole of its atoms = equivalent to that of the atomic mass Molar mass of a compound- the mass in grams of 1 mol of its moleucles or formula units = equivalent to the formula mass

Using the molar mass to count molecules by weighing; -

Using the example of CO2 from above Say we wanted to know the number of CO2 molecules in a sample of dry ice with a mass of 10.8 g 1 mol/44.01 g CO2 6.022 X 10^23 molecules/1 mol CO2 o Answer is 1.48 X10^23 molecules

Composition of compounds; -

A chemical formula in combination with the molar masses of its constituent elements indicates the relative quantities of each element in a compound One way to express how much of an element is in a given compound is to use the elements mass percent composition for that compound o The elements percentage of the compounds total mass o Formula;  Mass percent of element Y= mass of element Y in 1 mol of compound/ mass of 1 mol of the compound X 100% o Example; we want to calculate the mass percent composition of Cl in the chlorofluorocarbon CCL2 F2  2 X molar mass of Cl/ molar mass CCL2F2 X 100%  2 X 33.45 g/mol X 100%  120.91 g/mol  120.91 g/mol is the molar mass  58.64% is the answer

Mass percent composition as a conversion factor; -

Using the above example we found that the mass percent composition of CCL2F2 was 58.64%

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Which means there are 58.64 g Cl per hundred grams of CCL2F2 This can be expressed as a ratio; 58.64g Cl : 100g CCl2F2 Or it can be written as a fraction; 58.64g/100g CCl2F2 This can function as a conversion factor then

Conversion factors from chemical formulas; -

From the formula for CCl2F2 it tells us that 1 mol of it contains 2 mols of Cl atoms This can also be written as a ratio; 1 mol CCl2F2 : 2 mol Cl o Example; calculate the number of moles of Cl in 38.5 mol of CCl2F2  38.5 mol CCl2F2 X 2 mol Cl/ 1 mol CCl2F2  77.0 mol Cl

Determining a chemical formula from experimental data; -

Suppose in a lab they decomposed a sample of water and found that it produces 0.857g of hydrogen and 6.86g of oxygen How do you determine an empirical formula from this data? Empirical formula= represents a ratio of atoms or a ratio of moles to atoms First we have to convert everything from grams to moles o To convert to moles you have to divide each mass by the molar mass of the element o 0.857 gH X 1mol H/1.01 g H= 0.849 mol H o 6.86 gO X 1mol O/ 16.00 g O= 0.429 mol O o H0.849 O0.429  To get the smallest whole number subscripts in the formula we divide them by the smallest subscript present so in this case it would be 0.429  Which then leaves us with H1.98 O  AKA H2O when you round up

Determining molecular formulas for compounds; -

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Molecular formulas of a compound can be found from the empirical formula if we know the molar mass of the compound o Molecular formula is always a whole number multiple of the empirical formula Ex; find the molecular formula for fructose from its empirical formula CH2O and its molar mass is 180.2g/mol o Empirical formula molar mass= 12.01 g/mol + 2 (1.01 g/mol) +16.00 g/mol= 30.03 g/mol  n= 180.2/30.03= 6  this n value of 6 can now be used to find the molecular formula  (CH2O) X 6 = C6 H12 O6

Combustion analysis; -

The unknown compound undergoes combustion in the presence of pure oxygen When the sample burns all of the carbon converts to CO2 and all of the hydrogen converts to H2O The CO2 and H2O are then weighed We can then determine what the original amount of C and H were before combustion...