chemistry concept map 5 by

WEIGHING ATOMS + MOLECULES

ashlynhenderson2 — 2024-10-03 21:25:29 UTC

MOLAR MASS

ashlynhenderson2 — 2024-10-03 21:26:44 UTC

The molar mass of a substance represents the mass of 1 mole of said substance in grams. This value is expressed using g/mol.

MOLECULAR MASS

ashlynhenderson2 — 2024-10-03 21:27:20 UTC

The molecular mass of a substance is the sum of atomic masses of all atoms within a molecule/particle.

For example: H2O molecular mass = 2x Hydrogen atoms + 1x Oxygen atom = (2 * 1.008 amu) + (1 * 16.00 = 18.016 amu, allowing us to conclude 2 atoms of hydrogen and 1 atom of oxygen are equal to 1 mol of water.

FORMULA MASS

ashlynhenderson2 — 2024-10-03 21:27:57 UTC

The formula mass of a substance is the sum of atomic masses of all ions within an ionic compound.

For example: NaCl formula mass = 1x Sodium atom + 1x Chlorine atom = 22.99 amu + 34.35 amu = 58.44 amu, which essentially means that 1 mol of sodium and 1 mol of chlorine are equal to 1 mol of NaCl

THE MOLE

ashlynhenderson2 — 2024-10-03 21:28:57 UTC

Because atoms are so small, even the small amount of matter has a large amount of atoms, making it difficult for scientists to clearly express the number of atoms present within a certain substance. So, instead of expressing atomic weight with the SI unit for mass (g), chemists use the mole (mol), which represents this small quantity.

1 mole is equal to 6.02214 * 10^23 or Avogadro's number

MOLECULAR/FORMULA MASS CONVERSIONS

ashlynhenderson2 — 2024-10-03 21:30:02 UTC

grams to moles: grams (g) / molar mass (g/mol) -> moles (mol)

moles to grams: moles (mol) * molar mass (g/mol) -> grams (g)

particles to moles: # of particles / NA -> moles (mol)

moles to particles: moles (mol) * NA -> # of particles

MOLAR MASS CONVERSIONS

ashlynhenderson2 — 2024-10-03 21:30:22 UTC

grams to moles: grams (g) / molar mass (g/mol) -> moles (mol)

moles to grams: moles (mol) * molar mass (g/mol) -> grams (g)

moles to atoms: moles (mol) * NA -> # of atoms

atoms to moles: # of atoms / NA -> moles (mol)

MASS PERCENT COMPOSITION

ashlynhenderson2 — 2024-10-03 21:31:00 UTC

The mass % composition of a substance allows scientists to calculate the approximate mass contribution each element has towards the entire mass. This calculation is represented by:

( # of moles of element (n) * molar mass of element ) / molar mass of compound * 100%

which will equal the percentage of the element within this compound.

EMPIRICAL FORMULA

ashlynhenderson2 — 2024-10-03 21:32:23 UTC

As covered in previous chapters, a substances' empirical formula lists out elements within a compound by providing their proportions in a simple ratio rather than the actual amounts/arrangement. Previously, this formula was only attained by reducing molecular formulas, however, an empirical formula can also e collected from at mass percent composition.

ashlynhenderson2 — 2024-10-03 21:34:23 UTC

ashlynhenderson2 — 2024-10-03 21:35:07 UTC

MOLECULAR FORMULA

ashlynhenderson2 — 2024-10-03 21:39:33 UTC

A substances molecular formula lists out the elements found within an atom as well as the exact amount of atoms in each element present within that compound. This formula is typically provided or can be pieced together using the periodic table, however, this formula can also be derived from the compounds empirical formula.

AVOGADRO'S NUMBER

ashlynhenderson2 — 2024-10-03 21:59:14 UTC

Avogadro's number is a value used to represent 1 mol of any substance and it had no units.

For example, 1 inch = 2.54 cm and 6.02214*10^23 (atoms) = 1 mol (of atoms)

ashlynhenderson2 — 2024-10-03 22:00:17 UTC

ashlynhenderson2 — 2024-10-03 22:02:02 UTC

ashlynhenderson2 — 2024-10-03 22:10:08 UTC

ashlynhenderson2 — 2024-10-03 22:59:13 UTC

calculating mass % composition for H2O2:

first, identify the substances' molar mass:

-> 2 Hydrogen (2*1.008 g) + 2 Oxygen (2*16.00 g) = 34.016 g

then, calculate mass %

-> %H = 2 * 1.008g H / 34.016g H2O2 = 5.927%

-> %O = 2 * 16.00g O / 34.016g H2O2 = 94.073%

= 99.997 -> 100.00% (rounded hehe)

ashlynhenderson2 — 2024-10-03 23:26:51 UTC

calculating empirical formula from mass % composition:

(given 53.31% carbon, 11.19% hydrogen, and 35.51% oxygen)

first, convert % to grams by multiplying the percent value by the amount of compound available (100g in this case)

-> g C = 53.31% / 100g * 100g = 53.31 g C (53.31% = 53.31g C)

-> g H = 11.19% / 100g * 100g = 11.19 g H

-> g O = 35.51% / 100g * 100g = 35.51 g O

in each instance, the 100g compound cancels out with itself to leave just the new/same grams value

next, convert grams to moles by dividing each new mass value by its atomic mass

-> 53.31 g C * 1 mol / 12.01 g C = 4.439 moles of C

-> 11.19 g H * 1 mol / 1.008 g H = 11.10 moles of H

-> 35.51 g O * 1 mol / 16.00 g O = 2.219 moles of O

finally, reduce each new molar value to a ratio by dividing each result by the smallest value (smallest = 2.219 mol = oxygen)

-> C = 4.439 / 2.219 = 2

-> H = 11.10 / 2.219 = 5

-> O = 2.219 / 2.219 = 1

these values will be equal to the subscript for each element

final empirical formula: C2-H5-O

ashlynhenderson2 — 2024-10-03 23:40:50 UTC

calculating molecular formula from empirical formula:

(given CH empirical formula)

first, calculate the molar mass of the empirical formula for CH by adding their atomic masses

-> C = 12.01 g

-> H = 1.008 g

= 13.018 g

next, calculate the molar mass of CH -> (given = 78g)

next divide the molar mass by the empirical molar mass to get a molecular factor

-> 78g / 13.018g = 5.99

this value will be multiplied by the present subscripts to provide a new value equal to the molecular formula.

finally, multiply the current subscripts by the molecular factor

-> C: 1 subscript * 6.0 (5.99) = C6

-> H: 1 subscript * 6.0 (5.99) = H6

empirical formula: CH

molecular formula: C6H6

ashlynhenderson2 — 2024-10-04 00:11:00 UTC