Which of the following is not a correct formula unit? K 2 O MgO 2 Al 2 O 3 SrO

Welcome back everyone. We need to complete the following formulas by providing missing sub scripts for our given compounds. So right now we see for part one we have potassium bonded to an unknown subscript to our poly atomic ion which we recognize as sulfate, so recognize that this is a pol atomic ion because it's a charged ion with multiple atoms bonded to one another. We have sulfur bonded to three atoms of oxygen. And from memory we want to recall that cell fight Has a -3 an ion charge, meaning that when we recognize that potassium which is a metal is bonded here to our poly atomic ion. This combination yields a ionic compound meaning that we should recall. Ionic compounds involve a transfer of electric charge or of electrons. And so that minus three charge gets transferred to our potassium atom where potassium we recall is in group one A. Which is why our sulfide has a subscript of one which we which is implied we don't need to write it in. And so we would fill in that minus three charge as our missing subscript in the formula. So we would have K. three s. 0. three. And this would be our first answer for part one. Moving on to part two, we have S. R. With an unknown subscript bonded to our our foss fight poly atomic ion. So we have foss fight, Recall that foss fight has a -3 and ion charge and strontium Is considered based on its position on the periodic table. Being in group two a. It's considered a medal. And so this is the combination of yet again a medal plus a poly atomic ion phosphate meaning we have a transfer of electrons as we stated, forming our ionic compound. And so what we would have is the formation of strontium which will get as a subscript the charge from foss fight being three. And our foss fight which is going to get the charge of strontium. We recall again it's a group to a medal meaning it will form a plus two charge. And so foss fight is going to be P. +03 in parentheses with a subscript of two. And this would be our second answer for part two of the prompt. Now moving on to part three of our prompt. So there should be part three here we should recognize that we have yet again another polly atomic ion being in parentheses RSO three which we recognize as sulfite Recall that sulfide has a minus to an ion charge. Aluminum we want to recognize is also a medal. And on our periodic tables we find aluminum in group three a. Where we recall Adams in group three a form a plus three caddy in charge. And so yet again we have the combination of a medal and a poly atomic ion to form our ionic compound where our aluminum gets the charge of our sulfide an ion which is a minus to charge. So it has a subscript of two and our cell fight in parenthesis is going to get the charge of aluminum as a subscript of three. And this would be our third and final answer to complete this prompt as each of our missing sub scripts filled in for the corresponding formulas to the given compounds. I hope everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.