Question

Edited to make question more concise.

Can i use the outerWidth() value within the .css() method to offset the position of the “#item” by two times it’s width.

For example:

$("#item").css("left", - (2 x outerwidth(this)));

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The accepted answer, refers heavily to the original (more convoluted and broad) question I asked encompassing a wider set of functions I was developing.

elem.css('left', $(window).width()+'px'); 

This moved the element suitably, without relying on the outerWidth() value.

Answer 1

Something like this:

var elem = $("#item");

// 1. animate to left
elem.animate({'left': -(2 * elem.outerWidth())+'px'}, function() { 

    // 2. move to outside right border
    elem.css('left', $(window).width()+'px'); 

    // 3. animate to starting position 
    elem.animate({'left': '0px'});
});