Chapter 3: Mass, formula, and reactions

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/54

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 2:44 AM on 10/15/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

55 Terms

1
New cards

Counting term:

states specific number of items grouped together

2
New cards

Counting terms are used

when items are too small or there are too many to easily count one-by-one

3
New cards

24 cans =

1 case

4
New cards

144 pencils =

1 gross

5
New cards

500 sheets =

1 ream

6
New cards

12 eggs =

1 dozen

7
New cards

Moles are used to

count atoms, ions, and molecules (too small to count one-by-one)

8
New cards

1 mole =

6.02 x 10^23 atoms, molecules, or ions

9
New cards

Avogadro’s number =

6.02 x 10^23

10
New cards

Molar Mass (aka…

Molecular Weight, MW)

11
New cards

Molar Mass is

the mass in grams of 1 mole of an element or compound

12
New cards

1 mole C =

12.01 g

13
New cards

1 mole Li =

6.941 g

14
New cards

To find the molar mass of a compound:

Multiply the molar mass of each element by the subscript in the chemical formula then

add the masses.

15
New cards

Conversion factor to get from particles to mole:

1 mole/6.02 × 10^23 particles

16
New cards

Conversion factor to get from mole to particles:

6.02 × 10^23 particles/1 mole

17
New cards

Molar mass can be written as an equality and therefore as

conversion factors!

It converts grams to moles and vice versa.

18
New cards

Conversion factor to get from grams to mole:

ex. 16.05g CH4/1 mole of CH4

19
New cards

Conversion factor to get from mole to grams:

ex. 1 mole CH4/16.05g CH4

20
New cards

If your Given is atoms or molecules and your Need is moles use:

Avogadro’s #: 1 mole = 6.02 × 10^23 atoms or molecules

21
New cards

If your Given is moles and your Need is atoms or molecules use:

Avogadro’s #: 1 mole = 6.02 × 10^23 atoms or molecules

22
New cards

If your Given is grams and your Need is moles use:

Molar Mass: 1 mole = ____ grams

23
New cards

If your Given is moles and your Need is grams use:

Molar Mass: 1 mole = ____ grams

24
New cards

If your Given is atoms or molecules and your Need is grams use:

Avogadro’s # and Molar Mass

25
New cards

If your Given is grams and your Need is atoms or molecules use:

Avogadro’s # and Molar Mass

26
New cards

Chemical reaction:

transformation of one or more substances into different substances

27
New cards

Chemical equations:

use chemical formulas to express the identities of substances involved in a reaction and use coefficients to indicate their proportions

28
New cards

Reactants -

the substances that are consumed

29
New cards

Products -

the substances that are formed

30
New cards

Physical state of products/reactants:

(s) solid, (ℓ) liquid, (g) gas, (aq) aqueous solution

31
New cards

Coefficients:

indicate proportions of reactants and/or products

32
New cards

Stoichiometry:

- Quantitative relation between reactants and products in a chemical

equation

- Indicated in chemical equation by coefficients

33
New cards

Law of Conservation of Mass:

- The sum of the masses of the reactants in a chemical equation is equal to the sum of the masses of the products

34
New cards

Chemical equations must be balanced.

The number of atoms on the reactant side is equal to the number of atoms on the product side for each element.

35
New cards

To balance a chemical equation: Step 1

Write a chemical equation.

36
New cards

To balance a chemical equation: Step 2

Count the atoms of each element on the reactant side and product side

37
New cards

To balance a chemical equation: Step 3

Use ONLY coefficients to “balance.” You cannot change subscripts or chemical formulas. Balance elements that appear in only one reactant and product first

38
New cards

To balance a chemical equation: Step 4

Double check

39
New cards

Combustion Reactions:

heat-producing reactions between oxygen (O2) and another element or compound.

40
New cards

Combustion of hydrocarbons

(molecules composed of carbon and hydrogen) produces CO2 and H2O.

41
New cards

Rules for Balancing Combustion Reactions of Hydrocarbons: Step 1

Balance C first by adjusting coefficient in CO2

42
New cards

Rules for Balancing Combustion Reactions of Hydrocarbons: Step 2

Balance H next by adjusting coefficient for H2O

43
New cards

Rules for Balancing Combustion Reactions of Hydrocarbons: Step 3

Balance O last by adjusting coefficient for O2

44
New cards

Stoichiometric Calculations

To calculate the mass of a product from the mass of a reactant:

Step 1

Convert mass of reactant into moles using its molecular weight

45
New cards

Stoichiometric Calculations

To calculate the mass of a product from the mass of a reactant:

Step 2

Convert moles of reactant to moles of product using the mole ratio from the balanced chemical equation

46
New cards

Stoichiometric Calculations

To calculate the mass of a product from the mass of a reactant:

Step 3

Convert moles of product in mass of product using its molecular weight

47
New cards

Limiting reagent (aka limiting reactant limiting reactant):

reactant that is completely consumed in the chemical reaction and determine the amount of product that can be formed.

48
New cards

Theoretical Yield:

the maximum amount of product possible in a chemical reaction for given quantities of reactants.

Determined by the limiting reagent

49
New cards

Actual Yield:

the measured amount of product formed

50
New cards

Percent Yield: *hint* is a formula

= actual yield/theoretical yield x 100%

51
New cards

Percent Composition

Mass percent (%) = mass of element in compound/mass of compound x 100

52
New cards

We can calculate the empirical formula of a compound if we know the percent composition of each element.

1. Assume 100 g of substance.

2. Convert mass of each element to moles.

3. Compute mole ratios.

4. If necessary, convert to smallest whole-number ratios.

53
New cards

Empirical formula:

simplest whole-number molar ratio of elements in a compound

54
New cards

Molecular formula:

actual molar ratio of elements in a compound

• Equal to some multiple of empirical formula

• Determined from empirical formula and molecular mass

55
New cards