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Ira Lisetskai [31]
3 years ago
7

What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur

Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

                        Al                                               S

    1)            <u>  35.94  </u>    =1.33111                     <u>  64.06 </u>     = 2.001875

                       27                                               32

    2)          =    <u> 1.33111 </u>                                 = <u>  2.001875  </u>

                     1.33111                                         1.33111

                 

    3)         =   ( 1 ) × 2                                     =    ( 1.5 ) ×2

    4)         =  2                                                =   3

                     Empirical Formula = Al2S3

1) Divide the percentage given in question by the Relative Atomic Mass (RAM)

of the given elements.

2) When you find the answers of the first part of question, divide these once again but this time, by the lowest number you found in part 1.

3) and 4) Write down the values. If you get a decimal which is in between 0.3-0.7 (including the 0.3 and 0.7), you cannot make it a whole number by rounding of. Therefore, multiply the decimal with a whole number until you get a whole number as your answer. In this question, when you multiply 1.5 by 2, the answer is 3 which is a whole number. Multiply the other whole number by the same number as that you multiplied for 1.5. And use these numbers in part 4 to make the empirical formula which is Aluminium Sulfide (Al2S3)

Hope this helps you :)))

Please give this a brainliest

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<h3>What is Molarity?</h3>
  • The amount of a substance in a specific volume of solution is known as its molarity (M).
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Remember that mol/L is the unit of molarity (M).

We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.

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brainly.com/question/8732513

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