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

                        Al                                               S

    1)              35.94      =1.33111                      64.06     = 2.001875

                       27                                               32

    2)          =     1.33111                                 =  2.001875  

                     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

h let me think

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"