Chem 101 Ch. 6 – Flashcards

everything in nature is either chemically or physically combined with other substances to know the amount of a material in a sample, you need to know what fraction of the sample it is Some Applications: the amount of sodium in sodium chloride for diet the amount of iron in iron ore for steel production the amount of hydrogen in water for hydrogen fuel the amount of chlorine in freon to estimate ozone depletion

What if he bought a different size nail? Would the mass of a dozen be 0.150 lbs? Would there be 208 nails in 2.60 lbs? How would this effect the conversion factors?

If we can find the mass of a particular number of atoms, we can use this information to convert the mass of a element sample to the number of atoms in the sample. The number of atoms we will use is 6.022 x 1023 and we call this a mole 1 mole = 6.022 x 1023 things Like 1 dozen = 12 things

mole = number of particles equal to the number of atoms in 12 g of C-12 1 atom of C-12 weighs exactly 12 amu 1 mole of C-12 weighs exactly 12 g The number of particles in 1 mole is called Avogadro's Number = 6.0221421 x 1023 1 mole of C atoms weighs 12.01 g and has 6.022 x 1023 atoms the average mass of a C atom is 12.01 amu

The mass of one mole of atoms is called the molar mass The molar mass of an element, in grams, is numerically equal to the element's atomic mass, in amu

since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound

the relative weights of molecules can be calculated from atomic weights Formula Mass = 1 molecule of H2O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu since 1 mole of H2O contains 2 moles of H and 1 mole of O Molar Mass = 1 mole H2O = 2(1.01 g H) + 16.00 g O = 18.02 g

1 spider 8 legs 1 chair 4 legs 1 H2O molecule 2 H atoms 1 O atom

A molecular weight is the sum of the atomic weights of the atoms in a molecule. For the molecule ethane, C2H6, the molecular weight would be

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

So the percentage of carbon in ethane is...

the mass percent tells you the mass of a constituent element in 100 g of the compound the fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na this can be used as a conversion factor 100 g NaCl 39 g Na

Determine the mass of each element in 1 mole of the compound 2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g 1 mol O = 1(16.00 g) = 16.00 g Determine the molar mass of the compound by adding the masses of the elements 1 mole C2H5OH = 46.07 g

Divide the mass of each element by the molar mass of the compound and multiply by 100%

The simplest, whole-number ratio of atoms in a molecule is called the Empirical Formula can be determined from percent composition or combining masses The Molecular Formula is a multiple of the Empirical Formula

Hydrogen Peroxide Molecular Formula = H2O2 Empirical Formula = HO Benzene Molecular Formula = C6H6 Empirical Formula = CH Glucose Molecular Formula = c6h12o6 Empirical Formula = CH2O

Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O Assuming 100.00 g of para-aminobenzoic acid, C: 61.31 g x = 5.105 mol C H: 5.14 g x = 5.09 mol H N: 10.21 g x = 0.7288 mol N O: 23.33 g x = 1.456 mol O Calculate the mole ratio by dividing by the smallest number of moles: C: = 7.005 7 H: = 6.984 7 N: = 1.000 O: = 2.001 2 These are the subscripts for the empirical formula: c7h7no2

Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene? Solution: We are given an empirical formula and molecular weight and asked to determine a molecular formula. The subscripts in the molecular formula of a compound are whole-number multiples of the subscripts in its empirical formula. To find the appropriate multiple, we must compare the Molecular weight with the formula weight of the empirical formula, C3H4: 3(12.0 amu) + 4(1.0 amu) = 40.0 amu Next we divide the molecular weight by the empirical formula weight to obtain the multiple used to multiple the subscripts in C3H4: Molecular weight _______________________ = 121/40.0 = 3.02 Empirical formula weight Only whole number ratio make physical sense because we must be dealing with whole atoms. We therefore multiple each subscript in the empirical formula by 3 to give the Molecular formula C9H12.

The molecular formula is a multiple of the empirical formula To determine the molecular formula you need to know the empirical formula and the molar mass of the compound Molar Massreal formula Molar Massempirical formula= factor used to multiply subscripts

Determine the empirical formula May need to calculate it as previous C5H8 Determine the molar mass of the empirical formula 5 C = 60.05 g, 8 H = 8.064 g C5H8 = 68.11 g

Divide the given molar mass of the compound by the molar mass of the empirical formula Round to the nearest whole number 204g/68.11g=3

Multiply the empirical formula by the factor above to give the molecular formula (C5H8)3 = c15h24