Untitled Flashcards Set

 Moles, Mass, & Atoms

·        Molar mass is defined as the mass in grams of one mole of a substance. The Units of molar mass are grams per mole, abbreviated as g/mol. Molar Mas can be used to calculate the mass of a compound and to convert moles to grams and is the decimal number found on the periodic table (i.e. the average atomic mass.)

H  PO                                                             Al  (CO  )

^Do the same thing as the first problem.

Percent Composition:

·      Percent composition may be calculated from experimental data or from the atomic mass (molar mass) of the compound, using the following equation:

·      Percent (%) Comp. =

mass of element in compound/total mass of compound or molar mass x 100

1.    Calculate the percent composition of each element when it is found that 1.58 g Cu reacts with sulfur to form 1.98 g of the compound.

Cu= (1.58/1.98) x 100= 79.8% [made up of copper]

S=100-79.8= 20.2%

(0.40/1.98) x 100

2.    Determine the percent composition of the element that makes up magnesium phosphate.

Mg3(PO4)2 ---> Mg= 3 x 24.305= 72.915

                                    P= 2 x 30.974= 61.948

                                    O= 8 x 15.999= 127.992    +

                                                            --------------------

                                                            262.855 g/mol

Mg= 72.915/262.855= 27.7%

P= 61.948/262.855= 23.6%

O= 127.992/262.855= 48.7%

Stoichiometry:

·      Involves mass/mole relationships between reactants and products in a chemical reaction.

·      Stoichiometry is the areas of chemistry in which calculations of quantities in a chemical reaction are made.

                                              i.     You must have a balanced chemical equation.

                                            ii.     The coefficients in the equation are written in terms of moles. The relationship between these coefficients is called mole rations.

Example:

Sn (s) + 2 HF (g)  SnF  (s) + H  (g)

o   How many moles of hydrogen fluoride are needed to produce 2.5 moles of tin (2) fluoride

2.5 mole SnF  x (2mol HF/1mol SnF)= 5.0mol HF

o   What mass of tin is needed to produce 20.0 g of tine (2) fluoride?

20.0 g SnF  x (1mol/156.706g) (1mol Sn/ 1mol SnF  )

Divide the molar mass of the element/compound given to you

                                                                                                Mole Ratio

(118.71g/ 1mol)= 15.2 g Sn

X molar mass of what you are solving for

o   Phosphorous (P  ) burns in oxygen to form solid tetraphosphorous decoxide will be produced from 35.12 g of phosphorous?

P  + 5O   P  O

Why do reactions stop?

·      The limiting reactant limits the extent of the reaction and determines the amount of product formed.

·      Reactants leftover when a reaction stops are excess reactants.

Solving limiting reactant problems:

                        i.         Balance the chemical reaction given off if not already done.

                      ii.         Use stoichiometry for each individual reactant to find the mass of product produced.

a.    You will be provided a mass for each reactant.

b.   IF multiple products are present convert each reactant to THE SAME product.

                    iii.         The reactant that produces a lesser amount of product is the limiting reagent… the lesser amount of product is also the maximum amount that can be produced.

Example:

o   What mass of water can be made from 3.75gH  and 25.0gO

                        2H   (g)+O  (g)  2H  O(g)

The lesser answer is the right answer of the two

LR is the smaller one

ER is the larger one

o   If 35.4g CO and 11.5g H   , are combined to give the reaction below…

CO + 2H   CH   OH

a.    Identify the limiting reactant.

b.   How many grams of CH  OH could be produced?

LR=CO                       ER=H

c.    How much excess reactant remains?

…mass-to-mass conversion from the LR (original amount) to ER

35.4gCO(1mol/28.01molCO)(2molH,2/1molCO)(2.0158g/1mol)=5.10 H,2

[used]

11.5-5.10=6.40gH,2 [used]

o   Nitogen gas can be prepared by passing gaseous ammonia (18.1g) over solid copper (2) oxide (90.4g) at high temperatures

2 NH,3 + 3 CuO  _ N,2 + 3 Cu + 3 H,2O

%yield=(actual yield/ Theoretical Yield)x100

We will be given actual yield our answer is theoretical yield (the lesser answer)