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IrinaVladis [17]
3 years ago
15

One of the several oxides of tin found in the earthâs crust is 78.77% by mass tin. what is its empirical formula and name?

Chemistry
1 answer:
Zanzabum3 years ago
3 0
To determine the empirical formula of the compound given, we need to determine the ratio of each element in the compound. To do that we assume to have 100 grams sample of the compound with the given composition. Then, we calculate for the number of moles of each element. We do as follows:
         mass        moles
Sn     78.77       0.6636
O      21.23       1.3269

Dividing the number of moles of each element with the smallest value, we will have the empirical formula:

         moles                 ratio
Sn    0.6636          0.6636/0.6636 = 1
O     1.3269          1.3269/0.6636 = 2

The empirical formula would be SnO2.

You might be interested in
S8 + 24 F2 ⟶ 8 SF6
Arturiano [62]

Answer:

Theoretical Yield of SF₆ = 2.01 moles

Explanation: If you understand and can apply the methodology below, you will find it applies to ALL chemical reaction stoichiometry problems based on the balanced standard equation; i.e., balanced to smallest whole number coefficients.

Solution 1:

Rule => Convert given mass values to moles, solve problem using coefficient ratios. Finish by converting moles to the objective dimensions.

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

= 425g/256g/mol.      = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Determining Limiting Reactant => Divide moles each reactant by their respective coefficient; the smaller value will always be the limiting reactant. </em>

S₈ = 1.66/1 = 1.66

F₂ = 6.03/24 = 0.25 => F₂ is the limiting reactant

<em>Determining Theoretical Yield:</em>

Note: When working problem do not use the division ratio results for determining limiting reactant. Use the moles F₂ calculated from 229 grams F₂ => 6.03 moles F₂. The division procedure to define the smaller value and limiting reactant is just a quick way to find which reactant controls the extent of reaction.  

Given      S₈            +          24F₂            =>    8SF₆

             425g                    229g                      ?

   = 425g/256g/mol. = 226g/38g/mol.

= 1.66 moles S₈          = 6.03 moles F₂ <= Limiting Reactant

<em>Max #moles SF₆ produced from 6.03 moles F₂ and an excess S₈ </em>

Since coefficient values represent moles, the reaction ratio for the above reaction is 24 moles F₂ to 8 moles SF₆. Such implies that the moles of SF₆ (theoretical) calculated from 6.03 moles of F₂ must be a number less than the 6.03 moles F₂ given. This can be calculated by using a ratio of equation coefficients between 24F₂ and 8SF₆  to make the outcome smaller than 6.03. That is,

moles SF₆ = 8/24 x 6.03 moles = 2.01 moles SF₆ (=> theoretical yield)  

S₈ + 24F₂ => 8SF₆

moles SF₆ = 8/24(6.03) moles = 2.01 moles

You would NOT want to use 24/8(6.03) = 18.1 moles which is a value >> 6.03.        

This analysis works for all reaction stoichiometry problems.

Convert to moles => divide by coefficients for LR => solve by mole mole ratios from balanced reaction and moles of given.    

____________________

Here's another example just for grins ...

             C₂H₆O   +   3O₂     =>     2CO₂    + 3H₂O

Given:    253g          307g               ?               ?

a. Determine Limiting Reactant

b. Determine mass in grams of CO₂ & H₂O produced        

Limiting Reactant

moles  C₂H₆O = 253g/46g/mol = 5.5 moles  => 5.5/1 = 5.5

moles  O₂ = 307g/32g/mol = 9.6 moles         =><em>  9.6/24 = 0.4 ∴ O₂ is L.R.</em>

But the problem is worked using the mole values; NOT the number results used to ID the limiting reactant.  

 C₂H₆O   +       3O₂          =>     2CO₂    + 3H₂O

------------ 9.6 mole (L.R.)              ?               ?

mole yield CO₂ = 2/3(9.6)mole = 6.4 mole  (CO₂ coefficient < O₂ coefficient)

mole yield H₂O = 9.6mole  = 9.6mole (coefficients O₂ & CO₂ are same.)

mole used C₂H₆O = 1/3(9.6)mole = 3.2 mole (coefficient  C₂H₆O < coefficient O₂)

For grams => moles x formula weight (g/mole)

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