Stoichiometry: First, calculate the number of grams for one mole of Ca3 (PO3)4 (3 * (Mass of Ca)) + (4 * (Mass of P + (3 * Mass of Oxygen))) = (3*40.08) + 4(30.97 + (3*16.00)) =(120.24) + 4(78.97) =436.12 g / mol Ca3(PO3)4 This means there are 436.12 g per 1 mole of Ca(PO3)4. Since there are 4.50 moles of Calcium Phosphate, mulitply the molar mass of Ca(PO3)4 by 4.50 and you should get 1962.54 g. Since there are 3 sigfigs, the final answer is 1960 g.
on a side note: I put in all my work in case 1. your periodic table if different, 2. my work is wrong, 3. you put in the question wrong because I feel that the actual compound would be Ca3(PO4)3 instead of Ca3(PO3)4 (if this is the case, the answer should be 1820 g).
The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons. These opposite charges cancel each other out making the atom neutral.
